How does one prove the following integral? [math]\displaystyle\int_0^{\frac{\pi}2}\ln\left(\frac{\ln^2\sin heta}{\pi^2+\ln^2\sin heta} ight)\frac{\ln\cos heta}{ an heta}\,d heta=\frac{\pi^2}4[/math]

Question

How does one prove the following integral? [math]\displaystyle\int_0^{\frac{\pi}2}\ln\left(\frac{\ln^2\sin	heta}{\pi^2+\ln^2\sin	heta}ight)\frac{\ln\cos	heta}{	an	heta}\,d	heta=\frac{\pi^2}4[/math]

Marco Devincenzis · Accepted Answer

Here's a solution using complex analysis. As other people have shown, the integral is equivalent to:

[math]I=\frac{1}{2}\int_0^\infty\log\bigg(\frac{x^2}{x^2+\pi^2}\bigg)\log(1-e^{-2x})dx[/math]

I'll integrate the function:

[math]f(z)=\log z \log(1-e^{-2z})
[/math]

On this contour:

I'll spare you the gory details of showing that for [math]\epsilon \rightarrow 0[/math] and [math]R \rightarrow \infty[/math] the integrals on [math]\gamma_2[/math], [math]\gamma_4 [/math] and [math]\gamma_6 [/math] vanish. We are left with:

[math]\int_0^\infty \log x\log(1-e^{-2x})dx+[/math]
[math]\int_\infty^0\log(x+i\pi)\log(1-e^{-2x})dx+[/math][math]i\int_\pi^0\log z \log(1-e^{-2z})dz=0[/math]

I'll call the sum of the first two integrals [math]I_1[/math] and the last one [math]I_2[/math]. For [math]I_1 [/math]we have:

[math]I_1=\int_0^\infty \log\bigg(\frac{x}{\sqrt{x^2+\pi^2}}\bigg)\log(1-e^{-2x})dx-[/math]
[math]i\int_0^\infty \arg(x+i\pi)\log(1-e^{-2x})dx[/math]

Of which, we can see, the real part is very interesting. Now, for[math]I_2[/math] we have

[math]I_2=-i\int_0^\pi\log(iy)\log(1-e^{-2iy})dy=[/math]
[math]-i\int_0^\pi \bigg[\log y+i\frac{\pi}{2}\bigg][/math]
[math]\bigg[\log(2-2\cos 2y)+i \arg (1-e^{-2iy})\bigg]dy[/math]

Of which we are interested in the real part, which is:

[math]\Re I_2=\frac{\pi}{2}\int_0^\pi \log(2-2\cos 2y)dy + [/math]
[math]\int_0^\pi \log y \arg (1-e^{-2iy})dy=J_1+J_2[/math]

Let's evaluate the first one.

[math]J_1=\frac{\pi}{2}\int_0^\pi \log(2-2\cos 2y)dy=[/math]
[math]\frac{\pi}{2}\int_0^\pi \log(4\sin^2 y)dy=0[/math]

The last line follows from the well known integral of [math]\log(\sin x)[/math]. Let's evaluate [math]J_2[/math].

[math]1-e^{-2iy}=1-\cos (2y) + i \sin (2y)=[/math] 
[math]2 \sin^2 y + 2i \sin y \cos y=[/math]
[math]2 \sin y(\sin y +i \cos y)=2i \sin y e^{-iy}[/math]

We can now see that, up to unimportant costants, the multiplication by [math]2i\sin y[/math] is a rotation by [math]\pi/2[/math], so in terms of argument it's equivalent to a multiplication by [math]e^{i\pi/2}[/math]. So:

[math]\arg(1-e^{-2iy})=\arg (e^{i(\pi/2-y)})=\frac{\pi}{2}-y[/math]

So [math]J_2 [/math] becomes:

[math]J_2=\int_0^\pi \log y\frac{\pi}{2}dy-\int_0^{\pi}y\log y dy=-\frac{\pi^2}{4}[/math]

The last line follows from elementary integration techniques. Putting everything together we have:

[math]\Re I_1=-\Re I_2[/math]

[math]\int_0^\infty \log\bigg(\frac{x}{\sqrt{x^2+\pi^2}}\bigg)\log(1-e^{-2x})dx = \frac{\pi^2}{4}[/math]

So, multiplying by [math]2[/math] and remembering the original [math]1/2[/math] we have:

[math]I=\frac{\pi^2}{4}
[/math]

As desired.

Giordon Stark · Answer

[math]\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin	heta}{\pi^2+\ln^2\sin	heta}ight)\,\frac{\ln\cos	heta}{	an	heta}\,d	heta=\frac{\pi^2}{4}[/math]

Here's a stab at just trying to integrate. Almost all integration should be done using rudimentary knowledge. It doesn't quite look like the function is ill-defined on this range. You might think that [math]	heta = 0[/math] might cause a problem but
    [math]\lim_{	heta 	o 0} \frac{\ln \cos	heta}{	an	heta} = \lim_{	heta	o 0} \frac{-	an	heta}{\sec^2	heta} = 0[/math]

which isn't too surprising. It just seems like the function will go to zero on the limits, which is reasonable. So let's try to integrate. A first step is to imagine a right triangle where [math]\cos	heta = \frac{x}{h}[/math] and [math]\sin	heta = \frac{y}{h}[/math] and [math]h = \sqrt{x^2 + y^2}[/math] and initially do the substitution and see if things simplify (this rarely happens). In this case, it won't work.

The next step is to try and simplify using the Euler formula, and you'll have to be careful with the logarithm of complex (add in a phase, analytic continuation, branch cuts - which I won't deal with for now). IE:
    [math]z = r e^{i	heta} 	o w = \ln z = \ln r + i 	heta + 2n\pi i[/math] for [math]n = 0, \pm 1, \pm 2, ...[/math]

So what if we substitute terms like
    [math]\begin{align}
2\cos	heta &= e^{i	heta} + e^{-i	heta} = 	ext{Re}\left[e^{i	heta}ight]\
2i\sin	heta &= e^{i	heta} - e^{-i	heta} = 	ext{Im}\left[e^{i	heta}ight]
\end{align}[/math]
Again, doesn't seem to make our life easier. So let's try another substitution by letting [math]x = -\ln(\sin	heta)[/math] with [math]	ext{d}x = -\frac{	ext{d}	heta}{	an	heta}[/math]. This is done in Mike Wilkes [ https://www.quora.com/profile/Mike-Wilkes ]'s answer (correctly).
    [math]\frac{1}{2}\int_0^\infty\ln{\left(\frac{x^2}{x^2+\pi^2}ight)}\ln{(1-e^{-2x})}dx[/math]
Looking at this, it's a little bit easier. I'd like to refer to you a famous film: Stand and Deliver. Let's go ahead and integrate this! (or attempt to). Integrate by parts:
    [math]\frac{1}{2}\int_0^\infty \underbrace{\ln{\left(\overbrace{\frac{x^2}{x^2+\pi^2}}^{g}ight)}}_{	ext{d}v}\underbrace{\ln{(1-e^{-2x})}}_{u}dx[/math]
and this gets
    [math]\frac{1}{2} \left.\left[x \ln{g} + 2\pi \arctan\left(\frac{\pi}{x}ight)ight]\ln\left(1 - e^{-2x}ight) ight|_0^\infty[/math]
                      [math]- \int_0^\infty \left[x \ln{g} + 2\pi \arctan\left(\frac{\pi}{x}ight)ight] \frac{1}{e^{2x} - 1}\ 	ext{d}x[/math]
where the term [math]\left.uvight|_0^\infty 	o 0[/math] and you're left with the next integral which is somewhat easier to evaluate as I've gotten rid of at least one logarithm. EG: we are now down to solving
    [math]\int_0^\infty \left[x \ln{g} + 2\pi \arctan\left(\frac{\pi}{x}ight)ight] \frac{1}{1-e^{2x}}\ 	ext{d}x[/math]

I'm passing off the baton.

Mike Wilkes · Answer

I have not been able to "prove" that the integral equals [math]\frac{\pi^2}{4}[/math], but by the substitution [math]x=-\ln{(\sin{	heta})}[/math] (in which case the lower limit [math]	heta=0[/math] tends to [math]x=\infty[/math], while the upper limit [math]	heta=\pi/2[/math] tends to [math]x=0[/math]) the integral can be transformed to the simpler form:

[math]\frac{1}{2}\int_0^\infty\ln{\left(\frac{x^2}{x^2+\pi^2}ight)}\ln{(1-e^{-2x})}dx[/math]

This result requires also an interchange of the integration limits and a resulting sign change on the integral. I entered this, as suggested by S'leahcim Abarte [ https://www.quora.com/profile/Sleahcim-Abarte ], into WolframAlpha, integrating from 0 to 10, which gave the numerical approximation 2.4674, which is [math]\frac{\pi^2}{4}[/math] to four decimals.

This will take you to the URL of the WolframAlpha page containing the input and output:

integrate((1/2)*ln(y^2/(y^2+pi^2))*ln(1-e^(-2*y))dy from 0 to 10) [ http://www.wolframalpha.com/input/?i=integrate%28%281%2F2%29*ln%28y%5E2%2F%28y%5E2%2Bpi%5E2%29%29*ln%281-e%5E%28-2*y%29%29dy+from+0+to+10%29 ]

Brian Rost · Answer

I let Mathematica run overnight, and I have a fairly powerful machine, but it was unable to compute the closed form in that time. However, Mathematica also didn’t “give up” which it will sometimes do, returning the unevaluated integral back. So I cannot conclude that Mathematica can’t do it. I once had a very difficult integral I needed Mathematica to crunch that involved many complicated trigonometric functions, I let Mathematica run for about 18 hours before it gave me the solution. I then had to let it run another day to simplify the output. So maybe if you’re patient, Mathematica will eventually be able to provide a general closed form solution.

For the specific case [math]a=b=u=1[/math] mentioned by Emad Noujeim, Mathematica did return an exact result. It ran for approximately 14 minutes on my laptop.

[math]\frac{2 \pi \left(\sqrt{2 e \pi } \ \text{erf}\left(\frac{1}{\sqrt{2}}\right)-2\right)}{\sqrt{e}} \approx 3.13[/math]

Where [math]\text{erf}(x)[/math] is the error function: Error function - Wikipedia. Hope that helps!

For the specific case [math]a=b=u=1[/math] mentioned by Emad Noujeim, Mathematica did return an exact result. It ran for approximately 14 minutes on my laptop.

[math]\frac{2 \pi  \left(\sqrt{2 e \pi } \ 	ext{erf}\left(\frac{1}{\sqrt{2}}ight)-2ight)}{\sqrt{e}} \approx 3.13[/math]

Where [math]	ext{erf}(x)[/math] is the error function: Error function - Wikipedia [ https://en.wikipedia.org/wiki/Error_function ]. Hope that helps!

Toni Sementana · Answer

To solve this integral we use complex analysis. To do so, we set

[math]\displaystyle \ z=e^{ix} \longrightarrow \mathrm dx= \frac{1}{iz} \mathrm dz	ag*{}[/math]

Therefore,

[math]\displaystyle \begin{align}I_n&= \frac{1}{2^2} \oint \left( \frac{1}{2^n} \left(z+\frac{1}{z} ight)^n ight) \cdot \left( z^n + \frac{1}{z^n} ight) \frac{1}{i z} \mathrm dz \&= \frac{1}{i} \frac{1}{2^{n+2}} \oint \left( \frac{(z^2+1)^n}{z^n}ight) \cdot \left( \frac{z^{2n+1}}{z^n}ight) \cdot \frac{1}{z} \mathrm dz \ &= \frac{1}{i} \frac{1}{2^{n+2}} \oint \left( \frac{z^2+1}{z^2}ight)^n \cdot \left( z^{2n}+1ight) \cdot \frac{1}{z} \mathrm dz \&= \frac{1}{i} \frac{1}{2^{n+2}} \oint \left( \frac{1}{z} \left( z^2+1 ight)^n + \frac{1}{z} \left( \frac{z^2+1}{z^2} ight)^n ight) \mathrm dz \end{align}	ag*{}[/math]

Now for the integral

[math]\displaystyle \oint \left( \frac{1}{z} \left(z^2+1 ight)^n + \frac{1}{z} \left( \frac{z^2+1}{z^2} ight)^n ight) \mathrm dz 	ag*{}[/math]

we use Cauchy’s formula,

[math]\displaystyle\oint f(z)\mathrm dz = 2 \pi i \sum_{i} \mathcal{Res}(f(z);z_i)	ag*{}[/math]

Therefore we obtain

[math]\displaystyle I_n = \frac{\pi}{2^{n+1}} \left( 1+ 1ight) 	ag*{}[/math]

The final result is

[math]\displaystyle \boxed{\boxed{I_n= \frac{\pi}{2^n}}}	ag*{}[/math]

Lai Johnny · Answer

Latest Edit

After submitting the solution, I had generalised my result to even powers [math] 2m[/math] other than 6.

[math]\displaystyle I(m,n):=\int_{0}^{\frac{\pi}{4}}\left[\cos ^{2 m}(2 nx)+\sin ^{2 m}(2 n x)ight] \ln (1+	an x) d x= \frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}	ag*{} [/math]

where [math]m,n\in N. [/math]

Proof:

Letting [math] \displaystyle x\mapsto \frac{\pi}{4}-x [/math] yields

[math]\displaystyle \begin{aligned}I(m, n) &=\int_{0}^{\frac{\pi}{4}}\left[\sin ^{2 m}(2 x)+\cos ^{2 m}(2 x)ight) \ln \left(\frac{2}{1+	an x}ight]d x \&=\ln 2 \int_{0}^{\frac{\pi}{4}}\left[\sin^{2m} (2 n x)+\cos ^{2 m}(2 n x) ight] d x-I(m, n)\ I(m, n)&= \frac{\ln 2}{2} \int_{0}^{\frac{\pi}{4}}\left[\sin^{2m} (2n x)+\cos ^{2 m}(2nx)ight]d x\& = \frac{\ln 2}{4} \int_{0}^{\frac{\pi}{2}}\left[\sin^{2m} (n x)+\cos ^{2 m}(nx)ight]dx\&\stackrel{nx\mapsto x} {=}\frac{\ln 2}{2} \cdot \frac{1}{n} \int_{0}^{\frac{n \pi}{2}} \sin ^{2 m} x d x\&= \frac{\ln 2}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2 m} x d x\end{aligned}	ag*{} [/math]

Using Wallis Formula, we can conclude that [math]\displaystyle \boxed{I(m, n)=\frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}},	ag*{} [/math]

which is independent of[math]\displaystyle n [/math] .[math] [/math]

Back to our integral, when [math]m=3, [/math]

[math]\displaystyle I(3,2n)=\frac{5 \pi}{64} \ln 2 ,	ag*{} [/math]

[math]	extrm{ *******}	ag*{}[/math]

Original solution

When [math] n=0[/math],

[math] \displaystyle \int_{0}^{\frac{\pi}{4}} \ln (1+	an x) d x=\frac{\pi}{8} \ln 2 	ag*{} [/math]

When [math] n\in N[/math] , we first simplify

[math]\displaystyle \begin{aligned}\sin ^{6}(2 n x)+\cos ^{6}(2 n x) =& {\left[\sin ^{2}(2 n x)+\cos ^{2}(2 n x)ight]\left[\sin ^{4}(2 n x)-\sin ^{2}(2 n x) \cos ^{2}(2 n x)ight) } \&\left.+\cos ^{4}(2 n x)ight] \=& 1-3 \sin ^{2}(2 n x) \cos ^{2}(2 n x) \=& 1-\frac{3}{4} \sin ^{2}(4 n x) \=& 1-\frac{3}{8}(1-\cos 8 n x) \=& \frac{1}{8}(5+3 \cos (8nx))\end{aligned} 	ag*{} [/math]

To get rid of the natural logarithm, a simple substitution transforms the integral into

[math]\begin{aligned}I &=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln (1+	an x) d x \& \stackrel{x\mapsto\frac{\pi}{4}-x}{=} \frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(1+	an \left(\frac{\pi}{4}-xight)ight) d x \&=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(\frac{2}{1+	an x}ight) d x \&=\frac{1}{8} \ln 2 \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x) )d x-I \I &=\frac{\ln 2}{16} \int_{0}^{\frac{\pi}{4}}(5+3 \cos 8  n x) d x\&=\frac{\ln 2}{16}\left[5 x+\frac{3}{8 n} \sin (8 n x)ight]_0^{\frac{\pi}{4} }\ &=\frac{5 \pi}{64} \ln 2\end{aligned} 	ag*{} [/math]

Last but not least, how to deal with the odd one

[math]\displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(2 n+1) x+\cos ^{6}(2 n +1)xight] \ln (1+	an x) d x ?	ag*{} [/math]

Can you help?

Raziman T.V. (റസിമാൻ ടി.വി.) · Answer

This is just one of the many nails lying around, and there is one hammer to smash them all into oblivion: Euler's formula. [1]

Really, there are only two things you need. The formula itself:

[math]e^{ix} = \cos x + i \sin x 	ag{1}[/math]

And the fact that the integral of [math]e^{inx}[/math] vanishes from [math]0[/math] to [math]2\pi[/math] for integer [math]n
e 0[/math]. That is,

[math]\int\limits_0^{2\pi} e^{inx} dx = 2\pi\delta_{n0}	ag{2}[/math]

So let's hammer it in rightaway

[math]\sin 	heta = \displaystyle\frac{e^{i	heta}-e^{-i	heta}}{2i}[/math]

[math]\int\limits_0^{2\pi}\sin^{2n} 	heta d 	heta = \int\limits_0^{2\pi}d 	heta \left [\displaystyle\frac{1}{(2i)^{2n}}(e^{i	heta}-e^{-i	heta})^{2n}ight][/math]

Using binomial expansion,

[math]\int\limits_0^{2\pi}\sin^{2n} 	heta d 	heta = \int\limits_0^{2\pi}d 	heta \left [\displaystyle\frac{1}{(2i)^{2n}} \displaystyle\sum\limits_{k=0}^{2n} (-1)^k \binom{2n}{k} e^{i(2n-k)	heta}e^{-ik	heta}ight][/math]

Based on (2), we only need to care about the coefficient of [math]e^{i0	heta}[/math] which comes from [math]k=n[/math], giving us

[math]\int\limits_0^{2\pi}\sin^{2n} 	heta d 	heta = \int\limits_0^{2\pi}d 	heta \left [\displaystyle\frac{1}{(2i)^{2n}} (-1)^n \binom{2n}{n} ight][/math]

[math]= 2\pi\displaystyle\frac{\binom{2n}{n}}{2^{2n}}[/math]

And we are done.

More hammerings of nails aggregated here [ https://answeraggregator.quora.com/Eulers-formula-for-calculus ].

All right, so this one too has got autocollapsed due to too much latex. Quora really needs to get its act together. Someone is asking for a math proof, of course the answer is going to have latex in it!

1. Euler's formula - Wikipedia [ https://en.m.wikipedia.org/wiki/Euler%27s_formula ]

Stev Iones · Answer

For those of you who do not have access to Mathematica (very expensive), here’s how to do the double integration using the free open-source numerical computation package Octave [ https://www.gnu.org/software/octave/ ]:

[code]function f = f(x, y, a, b, u)
f = (b^2*cos(x).^2.*exp(-(u^2*cos(x).^2.*sin(y).^2)/2).*sin(y).^3)./(a^2*sin(x).^2.*sin(y).^2+b^2*cos(x).^2.*sin(y).^2+a^2*b^2*cos(y).^2);
endfunction

function g = g(a, b, u)
g = dblquad(@(x,y) f(x, y, a, b, u), 0, 2*pi, 0, pi);
endfunction

g(.5, .5, 3)
ans =  0.851955390903357

g(-1, .5, -4)
ans =  0.133605816072772
[/code]

Laxmi Narayan Bhandari · Answer

We have

[math]\mathcal{I} = \displaystyle \int_{\pi /4}^{\pi /2}\ln(\ln(	an{x}))\ \mathrm dx 	ag*{}[/math]

This is popularly known as the Vardi Integral.

Starting with susbtitution [math] 	an{x} \mapsto \dfrac{1}{x}[/math], we get

[math]\mathcal{I} = \displaystyle \int_0^1 \dfrac{\ln(-\ln{x})}{x^2+1}\ \mathrm dx	ag*{}[/math]

Using geometric series,

[math]\mathcal{I} = \displaystyle \sum_{k=0}^{\infty} (-1)^k \int_0^1 x^{2k}\ln(-\ln{x})\ \mathrm dx 	ag*{}[/math]

Using [math]x \mapsto e^{-x}[/math], we finally get

[math]\mathcal{I} = \displaystyle \sum_{k=0}^{\infty} (-1)^k\int_0^{\infty} e^{-(2k+1)x}\ln{x}\ \mathrm dx 	ag*{}[/math]

Let the final integral be [math]\mathcal{I_2}[/math].

To evaluate [math]\mathcal{I_2}[/math], consider the function [math]f[/math] with [math]a \in \R[/math],

[math] \begin{align} \displaystyle f(a) &= \int_0^{\infty} e^{-(2k+1)x} x^a \mathrm dx \ 	ext{using} \ (2k+1)x \mapsto x \ \displaystyle &= \dfrac{1}{(2k+1)^{a+1}}\int_0^{\infty} x^a e^{-x}\ \mathrm dx \ f(a)&= \dfrac{\Gamma(a+1)}{(2k+1)^{a+1}} \end{align}[/math]

Where [math]\Gamma(x)[/math] is the Gamma function defined as

[math] \displaystyle \Gamma(x)= \int_0^{\infty} t^{x-1}e^{-t}\ \mathrm dt 	ag*{}[/math]

Now, notice that

[math]\left. \dfrac{\partial}{\partial a}f(a) ight|_{a=0} =\mathcal{I_2}	ag*{}[/math]

Before continuing further, I would like to discuss about the Digamma function, [math]\psi(z)[/math] . The derivative of Gamma function is defined as

[math]\Gamma^{\prime}(z)= \Gamma(z)\psi(z)	ag*{}[/math]

Digamma function is defined as

[math]\displaystyle \psi(z+1)= -\gamma+\sum_{k=1}^{\infty} \dfrac{1}{k}-\dfrac{1}{k+z}	ag*{}[/math]

Here, [math]\gamma[/math] is the Euler–Mascheroni constant [ https://en.m.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant ].

Now,

[math]\dfrac{\partial}{\partial a}f(a)= \dfrac{\Gamma(a+1)(2k+1)^a [\psi(a+1)-\ln(2k+1) ]}{(2k+1)^{2a+1}}	ag*{}[/math]

So,

[math] \mathcal{I_2}= \dfrac{\Gamma(1)\psi(1)-\ln(2k+1)\Gamma(1)}{2k+1}  	ag*{}[/math]

Using the already discussed definitions for Gamma and Digamma functions,

[math]\Gamma(1)=1\ \psi(1)=-\gamma[/math]

Using them, we get the final expression

[math]\boxed{\boxed{ \mathcal{I_2}= \dfrac{-\gamma-\ln(2k+1)}{2k+1} }} 	ag*{}[/math]

Using this result,

[math]\displaystyle \mathcal{I} = -\gamma \sum_{k=0}^{\infty} \dfrac{(-1)^k}{2k+1}-\sum_{k=0}^{\infty} \dfrac{(-1)^k \ln(2k+1)}{2k+1}	ag*{}[/math]

Let

[math]\displaystyle S_1 = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{2k+1}\ \displaystyle S_2 = \sum_{k=0}^{\infty} \dfrac{(-1)^{k-1}\ln(2k+1)}{2k+1}[/math]

So,

[math]\mathcal{I_2}= -\gamma S_1 + S_2 	ag*{}[/math]

Before solving the sums, I would like to discuss about Dirichlet Beta function defined as

[math]\displaystyle \beta(s)=\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^s}	ag*{}[/math]

Differentiating with respect to [math]s[/math], the derivative of Dirichlet Beta function

[math]\displaystyle \beta^{\prime}(s)= \sum_{k=0}^{\infty} \dfrac{(-1)^{k-1}\ln(2k+1)}{(2k+1)^s}	ag*{}[/math]

Using these discussed functions and relations,

[math]S_1 = \beta(1) \ S_2 = \beta^{\prime}(1)[/math]

Using the series definition of [math]\arctan(1)[/math], you can easily prove that

[math]S_1=\beta(1)= \dfrac{\pi }{4}	ag*{}[/math]

I was struggling to find the value of [math]\beta^{\prime}(1)[/math] , so I checked the Wikipedia page of Dirichlet beta function  [ https://en.m.wikipedia.org/wiki/Dirichlet_beta_function ], and I got the required value.

[math]S_2=\beta^{\prime}(1)= \dfrac{\pi}{4} (\gamma-\ln{\pi})+\pi \ln \Gamma\left(\dfrac{3}{4}ight) 	ag*{}[/math]

Putting the values of the sums and simplifying, we get

[math]\mathcal{I}= \pi \ln \Gamma\left(\dfrac{3}{4}ight) -\dfrac{\pi}{4}\ln{\pi} 	ag*{}[/math]

Using the reflection formula for Gamma function

[math]\Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\sin(\pi x)}	ag*{}[/math]

And basic properties of logarithm, we get this beautiful result

[math]\boxed{\boxed{\pi \ln \left( \dfrac{ \sqrt{2 \pi \sqrt{\pi}}}{\Gamma(\frac{1}{4})}ight) }}	ag*{}[/math]

Brian Sittinger · Answer

We want to evaluate the definite integral

[math]I = \displaystyle \int_0^{\pi/2} \frac{\ln(\sin{x} + \cos{x})}{\sin{x}} \, dx. 	ag*{}[/math]

First of all, observe that

[math]\begin{align*} \displaystyle \int_0^{\pi/2} \frac{\cos{t} \cos{x}}{1 + \sin{t} \sin(2x)} \, dt &= \frac{\ln(1 + \sin{t} \sin(2x))}{2 \sin{x}} \Bigg|_{t = 0}^{t = \pi/2}\ &= \frac{\ln(1 + \sin(2x))}{2 \sin{x}}\ &= \frac{\ln((\sin{x} + \cos{x})^2)}{2 \sin{x}}\ &= \frac{\ln(\sin{x} + \cos{x})}{\sin{x}}. \end{align*} 	ag*{}[/math]

Using the result, we can express [math]I[/math] as a double integral. Then, we interchange the order of integration, obtaining

[math]I = \displaystyle \int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos{t} \cos{x}}{1 + \sin{t} \sin(2x)} \, dx \, dt. 	ag*{}[/math]

In order to evaluate the innermost integral, observe that [math](\sin{x} - \cos{x})^2 = 1 - \sin(2x)[/math], and then performing the interval-inverting substitution replacing [math]x[/math] with [math]\frac{\pi}{2} - x[/math] yields

[math]\begin{align*} \displaystyle \int_0^{\pi/2} \frac{\cos{t} \cos{x}}{1 + \sin{t} \sin(2x)} \, dx &= \int_0^{\pi/2} \frac{\cos{t} \cos{x} \, dx}{(1 + \sin{t}) - \sin{t} (\sin{x} - \cos{x})^2}\ &= \int_0^{\pi/2} \frac{\cos{t} \sin{x} \, dx}{(1 + \sin{t}) - \sin{t} (\sin{x} - \cos{x})^2}. \end{align*} 	ag*{}[/math]

Now, averaging the last two representations of the integral and letting [math]w = \sin{x} - \cos{x}[/math] gives us

[math]\begin{align*} \displaystyle \int_0^{\pi/2} \frac{\cos{t} \cos{x}}{1 + \sin{t} \sin(2x)} \, dx &= \frac{1}{2} \int_0^{\pi/2} \frac{\cos{t} (\cos{x} + \sin{x}) \, dx}{(1 + \sin{t}) - \sin{t} (\sin{x} - \cos{x})^2}\ &= \frac{1}{2} \int_{-1}^1 \frac{\cos{t}}{(1 + \sin{t}) - (\sin{t}) w^2} \, dw\ &= \cot{t} \cdot \int_0^1 \frac{1}{(1 + \csc{t}) - w^2} \, dw\ &= \frac{\cot{t}}{2\sqrt{1 + \csc{t}}} \ln\Big(\frac{\sqrt{1 + \csc{t}} + w}{\sqrt{1 + \csc{t}} - w}\Big) \Bigg|_{w=0}^{w=1}. \end{align*} 	ag*{}[/math]

Simplifying this further, we find that

[math]\begin{align*} \displaystyle \int_0^{\pi/2} \frac{\cos{t} \cos{x}}{1 + \sin{t} \sin(2x)} \, dx &= \frac{\cot{t}}{2\sqrt{1 + \csc{t}}} \ln\Big(\frac{\sqrt{1 + \csc{t}} + 1}{\sqrt{1 + \csc{t}} - 1}\Big)\ &= \frac{\cot{t}}{\sqrt{1 + \csc{t}}} \cdot \coth^{-1}{\sqrt{1 + \csc{t}}}. \end{align*} 	ag*{}[/math]

Hence, the double integral reduces to evaluating

[math]I = \displaystyle \int_0^{\pi/2} \frac{\cot{t}}{\sqrt{1 + \csc{t}}} \cdot \coth^{-1}{\sqrt{1 + \csc{t}}} \, dt. 	ag*{}[/math]

Making the substitution [math]z = \coth^{-1}{\sqrt{1 + \csc{t}}} = \frac{1}{2} \ln\Big(\frac{\sqrt{1 + \csc{t}} + 1}{\sqrt{1 + \csc{t}} - 1}\Big)[/math], we are essentially done with all of the difficult parts in evaluating this integral:

[math]I = \displaystyle \int_0^{\ln(1 + \sqrt{2})} 2z \, dz = \boxed{\ln^2(1 + \sqrt{2})}. \; \blacksquare 	ag*{}[/math]

Lai Johnny · Answer

[math]\quad \frac { \cos ^ { 4 } 	heta } { \cos ^ { 2 } \phi } + \frac { \sin ^ { 4 } 	heta } { \sin ^ { 2 } \phi } = 1 [/math]

[math]\Rightarrow ( \frac { \cos ^ { 2 } 	heta } { \cos \phi } ) ^ { 2 } + ( \frac { \sin ^ { 2 } 	heta } { \sin \phi } ) ^ { 2 } = 1[/math]

As [math]A( \frac { \cos ^ { 2 } 	heta } { \cos \phi }[/math],[math]\frac {\sin ^ { 2 } 	heta } { \sin \phi } )[/math] lies on a unit circle with origin as centre and its equation is[math] x^2+y^2=1.  [/math]

Since A[math]=[/math]([math]cos \alpha, sin \alpha), [/math]

[math]	extrm{therefore }\frac { \cos ^ { 2 } 	heta } { \cos \phi } = \cos \alpha 	extrm{ and } \frac { \sin ^ { 2 }	heta } { \sin \phi } = \sin \alpha \ 	extrm{ for some }\alpha \in R.[/math]

[math]\Rightarrow \cos ^ { 2 } 	heta = \cos \alpha \cos \phi  	extrm{ and } \sin ^ { 2 } 	heta = \sin \alpha \sin \phi [/math]

[math]\left. { \Rightarrow \cos \alpha \cos \phi + \sin \alpha \sin \phi = 1 } \ { \Rightarrow \cos ( \alpha - \phi ) = 1 } ight.[/math]

[math]\Rightarrow \alpha = \phi 	ext { or } 2 n \pi + \phi[/math]

Modified Version(suggested by Mr Shambhu Bhat )

[math]	extrm{Interchanging }\left. { 	heta 	extrm{ and } \phi 	extrm{ in the condition  gives}\ \frac { \cos ^ { 4 } \phi } { \cos ^ { 2 } 	heta } + \frac { \sin ^ { 4 } \phi } { \sin ^ { 2 } 	heta }  =1 } ight.[/math]

Original Version

[math]\Rightarrow \cos ^ { 2 } 	heta = \cos ^ { 2 } \phi  	extrm{ and } \sin ^ { 2 } 	heta = \sin ^ { 2 } \phi [/math]

[math]\left. { \quad \frac { \cos ^ { 4 } \phi } { \cos ^ { 2 } 	heta } + \frac { \sin ^ { 4 } \phi } { \sin ^ { 2 } 	heta } }\{ = \frac { \cos ^ { 4 } 	heta } { \cos ^ { 2 } 	heta } + \frac { \sin ^ { 4 } 	heta } { \sin ^ { 2 } 	heta } }\{ = \cos ^ { 2 } 	heta + \sin ^ { 2 } 	heta }\{ = 1 } ight.[/math]

Frank Wei · Answer

Well, since Roman Andronov [ https://www.quora.com/profile/Roman-Andronov ] has brilliantly detailed how to elegantly solve this integral, I’ll provide a less elegant way of integrating the expression using a combination of Euler’s formula, the binomial theorem, and a Fourier transform.

First off, let’s name [math]I[/math] to be our integral under question. Since the integrand is an event function, [math]f(x)=f(-x)[/math],

First off, let’s name [math]I[/math] to be our integral under question. Since the integrand is an event function, [math]f(x)=f(-x)[/math], we can rewrite our integral as

[math]\displaystyle I=\frac 12\int\limits_{-\pi/2}^{\pi/2}dx\,\cos^n x\cos nx	ag{1}[/math]

Next, we refer to an age-old identity using Euler’s formula

[math]\displaystyle\cos x=\frac {e^{ix}+e^{-ix}}2	ag{2}[/math]

Replacing [math]\cos^n x[/math] with (2) and taking out the extra factor of [math]2^n[/math] leaves us with

[math]\displaystyle I=\frac 1{2^{n+1}}\int\limits_{-\pi/2}^{\pi/2}dx\,\left(e^{ix}+e^{-ix}ight)^n\cos nx	ag{3}[/math]

All we have to do now is to show that the right-hand integral evaluates to one! This is easier said than done (it looks pretty intimidating, you have to admit). Fortunately, we can simplify it into a more recognizable form by using the trivial identity we all learned in first grade:

[math]\displaystyle (1+z)^n=\sum\limits_{r=0}^n\binom nr z^r	ag{4}[/math]

Factoring out [math]e^{-ix}[/math] from the parenthesis and replacing (3) with (4) gives us

[math]\displaystyle\begin{align*}I & =\frac 1{2^{n+1}}\int\limits_{-\pi/2}^{\pi/2}dx\,\left(1+e^{2ix}ight)^n e^{-nix}\cos nx\ & =\frac 1{2^{n+1}}\sum\limits_{r=0}^n\binom nr\int\limits_{-\pi/2}^{\pi/2}dx\, e^{ix(2r-n)}\cos nx\end{align*}	ag{5}[/math]

The resulting integral can be solved however you like. For this case, we can consider the function [math]f(x)[/math] which is defined as

[math]\displaystyle f(x)=\left\{\begin{array}{}e^{i\omega x}\qquad\quad	ext{for }-	frac {\pi}2%3Cx%3C	frac {\pi}2\\0\qquad\qquad	ext{otherwise}\end{array}ight.	ag*{}[/math]

Using the basic definition, the Fourier transform of [math]f(x)[/math] is equal to

[math]\displaystyle\begin{align*}\mathscr{F}\Bigr[f(x)\Bigr] & =\int\limits_{-\pi/2}^{\pi/2}dx\, e^{i\omega x}e^{-i\alpha x}\ & =\int\limits_{-\pi/2}^{\pi/2}dx\, \cos(x\omega-x\alpha)\ & =\frac {\sin(x\omega-x\alpha)}{\omega-\alpha}\,\Biggrvert_{-\pi/2}^{\pi/2}\ & =\frac {2\sin\bigr[	frac {\pi}2(\omega-\alpha)\bigr]}{\omega-\alpha}\end{align*}	ag*{}[/math]

where the imaginary portion equates to zero because sine is an odd function. Replacing the right-hand side of (5) with the result above, and canceling out the factors of two everywhere gives us a neat infinite sum

[math]\displaystyle\begin{align*}I & =\frac 1{2^{n+1}}\sum\limits_{r=0}^n\binom nr\left(\frac {\sin\left[\pi(r-n)ight]}{r-n}ight)\ & =\frac 1{2^{n+1}}\sum\limits_{r=0}^n\binom nr\left(\frac {\sin\left[\pi(n-r)ight]}{n-r}ight)\end{align*}	ag{6}[/math]

Now comes the tricky part: seeing that our little sum actually equals [math]\binom nn[/math]. To show this, we start off with the basic definition of the binomial coefficient and use Euler’s Reflective formula to get

[math]\displaystyle\binom yz=\frac {y!}{\Gamma(z+1)\Gamma(y-z+1)}=\frac {\sin\pi z}{\pi z}\frac {y!}{(y-z)\cdots(1-z)}	ag{7}[/math]

The next step is to use Melzak’s identity, which can be proven here [ https://arxiv.org/pdf/1603.07006.pdf ], into (7) and simplify to get

[math]\displaystyle\begin{align*}\binom yz & =\frac {\sin\pi z}{\pi z}\frac {y!}{(y-z)(y-z-1)\cdots(z-1)}\ & =\frac {\sin\pi z}{\pi z}\sum\limits_{r=0}^y\binom yr(-1)^r\frac {r}{z-r}\ & =\sum\limits_{r=0}^y\binom yr\frac {\sin\left[\pi(z-r)ight]}{\pi(z-r)}\end{align*}	ag*{}[/math]

Substitute the formula back into (6) to see that the finite sum equals [math]\binom nn[/math]. Therefore, it’s immediately established that

[math]\displaystyle\int\limits_{0}^{\pi/2}dx\,\cos^nx\cos nx=\frac {\pi}{2^{n+1}}	ag*{}[/math]

without using integration by parts!

Brian Sittinger · Answer

We are given the definite integral

[math]I = \displaystyle \int_0^{\pi/2} \frac{dx}{(1 + \sin{x})^2 \sqrt{\ln(	an{x} + \sec{x})}}. 	ag*{}[/math]

The integrand suggests using the substitution [math]t = \ln(\sec{x} + 	an{x})[/math] and [math]dt = \sec{x} \, dx[/math]. To use this, we first divide the numerator and denominator by [math]\cos^2{x}[/math]. Then, to deal with the remaining factor of [math]\sec{x}[/math], note that since [math]e^t = \sec{x} + 	an{x}[/math], we have [math]e^{-t} = \frac{1}{\sec{x} + 	an{x}} = \sec{x} - 	an{x}[/math], and thus [math]\sec{x} = \frac{1}{2}(e^t + e^{-t})[/math]. Then, this gives us

[math]\begin{align*} I &= \displaystyle \int_0^{\pi/2} \frac{\sec^2{x} \, dx}{(\sec{x} + 	an{x})^2 \sqrt{\ln(	an{x} + \sec{x})}}\ &= \int_0^{\infty} \frac{\frac{1}{2}(e^t + e^{-t}) \, dt}{(e^t)^2 \sqrt{t}}\ &= \frac{1}{2} \int_0^{\infty} (e^{-t} + e^{-3t})\, \frac{dt}{\sqrt{t}}. \end{align*} 	ag*{}[/math]

We are nearly there; now we make the rationalizing substitution [math]w = \sqrt{t}[/math]:

[math]\begin{align*} I &= \displaystyle  \frac{1}{2} \int_0^{\infty} (e^{-w^2} + e^{-3w^2}) \cdot 2 \, dw\ &= \int_0^{\infty} e^{-w^2} \, dw + \int_0^{\infty} e^{-3w^2} \, dw. \end{align*} 	ag*{}[/math]

For the latter integral, we use the rescaling substitution [math]y = w\sqrt{3}[/math]. Then, we can use the classic Gaussian integral to complete the evaluation of [math]I[/math]:

[math]\begin{align*} I &= \displaystyle \int_0^{\infty} e^{-w^2} \, dw + \int_0^{\infty} e^{-y^2} \cdot \Big(\frac{1}{\sqrt{3}} \, dy\Big)\ &= \Big(1 + \frac{1}{\sqrt{3}}\Big) \int_0^{\infty} e^{-w^2} \, dw\ &= \boxed{\frac{\sqrt{\pi}}{2} \Big(1 + \frac{1}{\sqrt{3}}\Big)}. \end{align*} 	ag*{}[/math]

Brian Sittinger · Answer

We want to evaluate the integral

[math]I = \displaystyle \int_0^{\pi/2}\ln^2(\cos{x}) \, dx. 	ag*{}[/math]

In the solution below, I give a derivation that does not use Fourier series. It is a bit on the long side.

To this end, we first observe that by using the interval-inverting substitution replacing [math]x[/math] with [math]\frac{\pi}{2} - x[/math] along with the cofunction identity that

[math]I = \displaystyle \int_0^{\pi/2}\ln^2(\sin{x}) \, dx. 	ag*{}[/math]

Therefore, we see that

[math]\begin{align*} I &= \displaystyle \frac{1}{2} \int_0^{\pi/2} \Big(\ln^2(\cos{x}) + \ln^2(\sin{x})\Big) \, dx\ &= \frac{1}{4} \int_0^{\pi/2} \Big[\Big(\ln(\sin{x}) + \ln(\cos{x})\Big)^2 + \Big(\ln(\sin{x}) - \ln(\cos{x})\Big)^2\Big] \, dx\ &= \frac{1}{4} \int_0^{\pi/2} \ln^2(\sin{x} \cos{x})  \, dx + \frac{1}{4} \int_0^{\pi/2} \ln^2(	an{x}) \, dx. \end{align*} 	ag*{}[/math]

Denoting these latter integrals as [math]J[/math] and [math]K[/math], respectively, we now evaluate these one at a time.

[math]\begin{align*} J &= \displaystyle \int_0^{\pi/2} \ln^2\Big(\frac{1}{2} \sin(2x)\Big) \, dx\ &= \frac{1}{2} \int_0^{\pi} \ln^2\Big(\frac{1}{2} \sin{t}\Big) \, dt, 	ext{ via } t = 2x\ &= \int_0^{\pi/2} \ln^2\Big(\frac{1}{2} \sin{t}\Big) \, dt, 	ext{ via symmetry of the integrand}\ &= \int_0^{\pi/2} \Big(\ln^2{2} - 2 \ln{2} \cdot \ln(\sin{t}) + \ln^2(\sin{t})\Big) \, dt\ &= \frac{\pi}{2} \ln^2{2} - 2 \ln{2} \cdot \Big(-\frac{\pi}{2} \ln{2}\Big) + I\ &= \frac{3\pi}{2} \ln^2{2} + I. \end{align*} 	ag*{}[/math]

As for the second integral,

[math]\begin{align*} K &= \displaystyle \int_0^{\pi/2} \ln^2(	an{x}) \, dx\ &= \int_0^{\infty} \frac{\ln^2{t}}{t^2 + 1} \, dt, 	ext{ via } t = 	an{x}\ &= \int_0^1 \frac{\ln^2{t}}{t^2 + 1} \, dt + \int_1^{\infty} \frac{\ln^2{t}}{t^2 + 1} \, dt\ &= \int_1^{\infty} \frac{\ln^2{t}}{t^2 + 1} \, dt + \int_1^{\infty} \frac{\ln^2{t}}{t^2 + 1} \, dt, 	ext{ letting } t 	o \frac{1}{t} 	ext{ in first integral}\ &= 2\int_1^{\infty} \frac{\ln^2{t}}{t^2 + 1} \, dt\ &= 2 \int_0^{\infty} \frac{w^2}{e^{-2w} + 1} \cdot e^{-w} \, dw, 	ext{ via } t = e^{-w}. \end{align*} 	ag*{}[/math]

Next, we use the geometric series:

[math]\begin{align*} K &= \displaystyle 2 \int_0^{\infty} w^2 e^{-w} \cdot \sum_{n=0}^{\infty} (-e^{-2w})^n \, dw\ &= 2 \sum_{n=0}^{\infty} (-1)^n \int_0^{\infty} w^2 e^{-(2n+1)w} \, dw\ &= 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3} \int_0^{\infty} u^2 e^{-u} \, du, 	ext{ via } u = (2n+1)w\ &= 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3} \cdot 2!\ &= 4 \cdot \frac{\pi^3}{32}\  &= \frac{\pi^3}{8}. \end{align*} 	ag*{}[/math]

A proof to the series result used in the last two lines can be found here:

Brian Sittinger's answer to Can one evaluate  [ https://www.quora.com/Can-one-evaluate-displaystyle-sum_-k-0-infty-frac-1-k-2k-1-3-alternatively-without-using-the-concept-of-Fourier-Series/answer/Brian-Sittinger ][math] \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^3} [/math] alternatively without using the concept of Fourier Series? [ https://www.quora.com/Can-one-evaluate-displaystyle-sum_-k-0-infty-frac-1-k-2k-1-3-alternatively-without-using-the-concept-of-Fourier-Series/answer/Brian-Sittinger ]

Putting this all together, we see that

[math]I = \displaystyle \frac{1}{4} \Big(\frac{3\pi}{2} \ln^2{2} + I\Big) + \frac{1}{4} \cdot \frac{\pi^3}{8}. 	ag*{}[/math]

Solving for [math]I[/math], we conclude that

[math]\displaystyle I = \int_0^{\pi/2}\ln^2(\cos{x}) \, dx = \frac{\pi}{2} \ln^2{2} + \frac{\pi^3}{24} \approx 2.04662. 	ag*{}[/math]

Roman Andronov · Answer

What, most likely, is expected of you (by your Real Analysis professor (or teacher)) is to show that integrating:

[math]\displaystyle C_n = \int_0^{\frac{\pi}{2}}\cos^nx\cdot \cos nx \; dx 	ag{1}[/math]

by parts once (assuming that [math]n\in \mathbb{N}[/math]) we obtain:

[math]\displaystyle C_n = \int_0^{\frac{\pi}{2}}\cos^{n-1}x\cdot \sin nx \cdot \sin x \; dx 	ag{2}[/math]

(make sure you understand how (2) came into existence)

Add one copy of [math]C_n[/math] to both sides of (2):

[math]\displaystyle 2\cdot C_n = \int_0^{\frac{\pi}{2}}\left(\cos^{n-1}x\cdot  \sin nx\cdot  \sin x + \cos^nx\cdot  \cos nxight) \; dx 	ag{3}[/math]

Transform the integrand in (3) by factoring out one copy of [math]\cos^{n-1}x[/math] from it:

[math]\cos^{n-1}x\cdot \sin nx\cdot  \sin x + \cos^nx \cdot \cos nx = 	ag*{}[/math]

[math]\cos^{n-1}x(\sin nx\cdot  \sin x + \cos x\cdot  \cos nx) 	ag{4}[/math]

Isn’t it beautiful: we have an expression in (4) that collapses into a cosine of a difference due to:

[math]\cos(x-y) = \cos x\cdot  \cos y + \sin x\cdot  \sin y 	ag*{}[/math]

identity, from where:

[math]\cos^{n-1}x(\sin nx\cdot  \sin x + \cos x\cdot  \cos nx) = \cos^{n-1}x\cos(n-1)x 	ag{5}[/math]

Put (5) into (3):

[math]\displaystyle 2\cdot C_n = \int_0^{\frac{\pi}{2}}\cos^{n-1}x\cdot  \cos ((n-1)x) \; dx 	ag{6}[/math]

But the integral on the RHS of (6) can be expressed in terms of the original integral as follows:

[math]2\cdot C_n = C_{n-1} 	ag{7}[/math]

which represents a certain recurrence relation. We can bring the full might of the generating functions apparatus to bare on (7) - as a valid and interesting combination of problem-solving techniques - but I hope that in this particular case there is no need to do so:

[math]2^2C_n = C_{n-2} 	ag*{}[/math]

[math]2^3C_n = C_{n-3} 	ag*{}[/math]

and so on until:

[math]2^nC_n = C_0 	ag{8}[/math]

But:

[math]\displaystyle C_0 = \int_0^{\frac{\pi}{2}}\cos^0x\cdot  \cos 0 \; dx = \dfrac{\pi}{2} 	ag{9}[/math]

Put (9) into (8):

[math]2^n\cdot C_n = \dfrac{\pi}{2} 	ag{10}[/math]

Solve (10) for [math]C_n[/math]:

[math]\displaystyle C_n = \int_0^{\frac{\pi}{2}}\cos^nx\cdot \cos nx \; dx = \dfrac{\pi}{2^{n+1}} 	ag{10}[/math]

(thinking out loud: other ideas to kick around are: integration in a complex plane and a Fourier transform (in particular - the cosine decomposition))

Extra for experts.

If you think that you’ve got the hang of the basic idea, see if you can prove that:

[math]\displaystyle S_n = \int_0^{\frac{\pi}{2}}\cos^nx\cdot \sin nx \; dx = \dfrac{1}{2^{n+1}}\sum_{k=1}^n\dfrac{2^k}{k} 	ag{12}[/math]

For more information on and ideas about problem-solving in mathematics, physics and computer science please visit my YouTube channel ProbLemma [ https://www.youtube.com/@ProbLemmaChannel/videos ].

Enrico Gregorio · Answer

It doesn’t hold.

What’s true is that, in a neighborhood of [math]0[/math],

[math]\dfrac{2\sin	heta	an	heta+\sin	heta	an^2	heta}{	heta^2}=\dfrac{2(	heta^2+o(	heta^2))+o(	heta^2)}{	heta^2}[/math]

because

[math]\sin	heta=	heta+o(	heta)[/math]

[math]	an	heta=	heta+o(	heta)[/math]

Thus we can simplify to

[math]2+o(1)[/math]

and the limit for [math]	heta	o0[/math] is [math]2.[/math]

Alexey Godin · Answer

Take [math]x=\frac{\pi}{4}-t[/math]

Note that [math]\sin^6(2nx)+\cos^6(2nx)=\sin^6(2nt)+\cos^6(2nt)[/math] (consider [math]n=4k, 4k\pm 1,4k+2[/math])

So our integral

[math]\displaystyle I=\int_0^\frac{\pi}{4} (\sin^6(2nt)+\cos^6(2nt))\ln(1+\frac{1-	an t}{1+	an t})dt[/math]

Now sum both of integrals to get rid of the logarithm:

[math]\displaystyle 2I=\int_0^\frac{\pi}{4}(\sin^6(2nx)+\cos^6(2nx))\ln 2 dx= [/math]

[math]\displaystyle \ln 2\int_0^\frac{\pi}{4} \sin^4(2nx)+\cos^4(2nx)-\frac{1}{4}\sin^2(4nx) dx =[/math]

[math]\displaystyle\ln 2\int_0^\frac{\pi}{4} 1-\frac{3}{4}\sin^2(4nx) dx =[/math]

[math]\displaystyle \ln 2\int_0^\frac{\pi}{4} \frac{5}{8}-\frac{3}{8}\cos(8nx) dx=\frac{5\pi\ln 2}{32}[/math]

Thus

[math]I=\frac{5\pi\ln 2}{64}[/math]

Ragu Rajagopalan · Answer

[math]\[/math]

[math]I = \displaystyle\int_0^{2\pi} \sqrt{\cos^8 x \cdot \sin^2 x + \sin^8x \cdot \cos^2x} \cdot \mathrm dx [/math]

[math]\qquad = \displaystyle\int_0^{2\pi} \sqrt{\cos^6 x + \sin^6x} \cdot \sin x \cdot \cos x \cdot \mathrm dx [/math]

[math]\qquad = \displaystyle\int_0^{2\pi} \sqrt{\left(\dfrac{1 + \cos 2x}{2}\right)^3 + \left(\dfrac{1 - \cos 2x}{2}\right)^3} \cdot  \dfrac{\sin 2x}{2} \cdot \mathrm dx [/math]

[math]\qquad = \dfrac{1}{4} \cdot \displaystyle\int_0^{2\pi} \sqrt{1 + 3\cos^2 2x} \cdot  \sin 2x \cdot \mathrm dx [/math]

[math]\text{Splitting the range into 4 regions from 0 to } \: \dfrac{\pi}{2}  \: : [/math]

[math]\qquad = \dfrac{1}{4} \cdot 4 \cdot \displaystyle\int_0^{\frac{\pi}{2}} \sqrt{1 + 3\cos^2 2x} \cdot  \sin 2x \cdot \mathrm dx [/math]

[math]\text{Let } \: \tan y = \sqrt{3}\cdot \cos 2x \implies \sin 2x \cdot \mathrm dx = \dfrac{(-1) \cdot \sec^2 y \cdot \mathrm dy}{2\sqrt{3}} [/math]

[math]\text{New limits would be : }\: \dfrac{\pi}{6} \: \text{ to } \: \dfrac{-\pi}{6} [/math]

[math]\therefore I = \displaystyle\int_{\frac{\pi}{6}}^{\frac{-\pi}{6}} \sqrt{1 + \tan^2 y} \cdot  \dfrac{(-1) \cdot \sec^2 y \cdot \mathrm dy}{2\sqrt{3}} [/math]

[math]\qquad = \dfrac{1}{2\sqrt{3}} \cdot \displaystyle\int_{\frac{\pi}{6}}^{\frac{-\pi}{6}} \sec^3 y \cdot \mathrm dy [/math]

[math]\qquad = \dfrac{1}{2\sqrt{3}} \cdot \dfrac{1}{2} \cdot \left[\sec y \cdot \tan y + \ln \left|\sec y + \tan y\right| \:\right]_{\frac{\pi}{6}}^{\frac{-\pi}{6}} [/math]

[math]\qquad = \dfrac{1}{4\sqrt{3}} \cdot \left[\left(2\sqrt{3} + \ln (2 + \sqrt{3}) \right) - \left(-2\sqrt{3} + \underbrace{\ln(2-\sqrt{3})}_{=\: -\ln(2+\sqrt{3})} \right)\right] [/math]

[math]\qquad = \dfrac{2}{4\sqrt{3}} \cdot \left[2\sqrt{3} + \ln (2 + \sqrt{3}) \right] = 1 + \dfrac{\ln(2+\sqrt{3})}{2\sqrt{3}} = \text{RHS}[/math]

PROVED

Graph from Desmos :

Brian Sittinger · Answer

Use integration by parts where [math]u = \ln(\sin{x})[/math] and [math]dv = \sin{x} \, dx[/math]:

[math]\begin{align*} \displaystyle \int \sin{x} \ln(\sin{x}) \, dx &= \ln(\sin{x}) \cdot(-\cos{x}) - \int \frac{\cos{x}}{\sin{x}} \cdot (-\cos{x}) \, dx\ &= \displaystyle -\cos{x} \ln(\sin{x}) + \int \frac{\cos^2{x}}{\sin{x}}  \, dx. \end{align*} 	ag*{}[/math]

Next, we use the classic pythagorean trigonometric identity to finish the antidifferentiation.

[math]\begin{align*} \displaystyle \int \sin{x} \ln(\sin{x}) \, dx &=  \displaystyle -\cos{x} \ln(\sin{x}) + \int \frac{1 - \sin^2{x}}{\sin{x}} \, dx\ &= \displaystyle -\cos{x} \ln(\sin{x}) + \int (\csc{x} - \sin{x}) \, dx\ &= \displaystyle -\cos{x} \ln(\sin{x}) - \ln|\csc{x} + \cot{x}| + \cos{x} + C. \end{align*} 	ag*{}[/math]

Why did I do the antidifferentiation first? The integral in question is improper at 0, and we have to be careful taking the limit at 0.

[math]\begin{align*} I &= \displaystyle \int_0^{\pi/2} \sin{x} \ln(\sin{x}) \, dx \ &= \displaystyle \lim_{t 	o 0^+} \Big(-\cos{x} \ln(\sin{x}) - \ln(\csc{x} + \cot{x}) + \cos{x} \Big) \Big|_t^{\pi/2} \ &= \displaystyle 0 - \lim_{t 	o 0^+} \Big(-\cos{t} \ln(\sin{t}) - \ln(\csc{t} + \cot{t}) + \cos{t} \Big)\ &= \displaystyle \lim_{t 	o 0^+} \Big(\cos{t} \ln(\sin{t}) + \ln(\csc{t} + \cot{t}) - \cos{t} \Big)\. \end{align*} 	ag*{}[/math]

All that remains is carefully computing the limit as [math]t 	o 0^+[/math], which we accomplish by trigonometric identity manipulations:

[math]\begin{align*} I &= \displaystyle \lim_{t 	o 0^+} \Big(\cos{t} \ln(\sin{t}) + (\ln(1 + \cos{t}) - \ln(\sin{t})) - \cos{t}\Big)\ &= \displaystyle (\ln{2} - 1) + \lim_{t 	o 0^+} (\cos{t} - 1) \ln(\sin{t}) \&= \displaystyle (\ln{2} - 1) + \lim_{t 	o 0^+} (\cos{t} - 1) \ln(\sin{t}) \cdot \frac{\cos{t} + 1}{\cos{t} + 1}\ &= \displaystyle (\ln{2} - 1) - \lim_{t 	o 0^+} \ln(\sin{t}) \cdot \frac{\sin^2{t}}{\cos{t} + 1}\&= \displaystyle (\ln{2} - 1) - \frac{1}{2} \lim_{t 	o 0^+} \frac{\ln(\sin{t}) }{\csc^2{t}}\end{align*} 	ag*{}[/math]

Now, we can finish the limit (and the integral) by a simple application of L’Hopital’s Rule:

[math]\begin{align*} I &= \displaystyle (\ln{2} - 1) - \frac{1}{2} \lim_{t 	o 0^+} \frac{\cot{t}}{-2\csc{t} \cdot \csc{t} \cot{t}}\ &= \displaystyle (\ln{2} - 1) +  \lim_{t 	o 0^+} \sin^2{t}\ &= \ln{2} - 1. \end{align*} 	ag*{}[/math]