# twist on classic log of sine and cosine integral

I ran across this integral and have not been able to evaluate it.

$\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\ln(\cos(x))dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$

I had some ideas. Perhaps some how arrive at $\displaystyle\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{4}}=\frac{{\pi}^{4}}{192}$.

and $\displaystyle \ln(2)\int_{0}^{\frac{\pi}{2}}x\ln(2)dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}$

by using the identity $\displaystyle\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}=-x\ln(\sin(x))-x\ln(2)$

and/or $\displaystyle \ln(\cos(x))=-\ln(2)-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}$

I have used the first one to evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))dx$, so I thought perhaps it could be used in some manner here.

I see some familiar things in the solution, but how to get there?.

Does anyone have any clever ideas?.

Thanks.

-

$$\zeta(4):=\sum_{n=1}^\infty\frac{1}{n^4}=\frac{\pi^4}{90}\Longrightarrow \zeta_2(4):=\sum_{n=1}^\infty\frac{1}{(2n)^4}=\frac{1}{16}\zeta(4)=\frac{\pi^4}{16\cdot 90}\Longrightarrow$$

$$\Longrightarrow\sum_{n=0}^\infty\frac{1}{(2n+1)^4}=\zeta(4)-\zeta_2(4)=\frac{15}{16}\frac{\pi^4}{90}=\frac{\pi^4}{96}$$

-
Thanks, Don. I think I managed to get it figured out. I used the series I mentioned in my first post. – Cody Oct 5 '12 at 21:19

I thought I would come back and show what I done. I am rather uneasy about this solution and you'll see why.

Using the identities mentioned previously:

$\displaystyle -\ln(\sin(x))=\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}+\ln(2)$

and $\displaystyle -\ln(\cos(x))=\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}+\ln(2)$

I subbed them in and arrived at:

$\displaystyle\int_{0}^{\frac{\pi}{2}}\left(\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}+x\ln(2)\right)\left(\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}+\ln(2)\right)dx$

$=\displaystyle \int_{0}^{\frac{\pi}{2}}(\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}$ +$\displaystyle \ln(2)\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}+\ln(2)\sum_{k=1}^{\infty}\frac{(-1)^{k}x\cos(2kx)}{k}+x\ln^{2}(2))dx$

Now, here I made an otherwise 'illegal' move. I took the product of the sums under one summation.

$\displaystyle\int_{0}^{\frac{\pi}{2}}(\sum_{k=1}^{\infty}\frac{(-1)^{k}x\cos^{2}(2kx)}{k^{2}}$ $+\displaystyle\ln(2)\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}$ $+\displaystyle\ln(2)\sum_{k=1}^{\infty}\frac{(-1)^{k}x\cos(2kx)}{k}$ $+x\ln^{2}(2))dx$

Switch the sum and integral:

$\displaystyle\sum_{k=1}^{\infty}(\underbrace{\int_{0}^{\frac{\pi}{2}}\frac{(-1)^{k}x\cos^{2}(2kx)}{k^{2}}dx}_{\text{[1]}}$ $+\displaystyle\underbrace{\ln(2)\int_{0}^{\frac{\pi}{2}}\frac{x\cos(2kx)}{k}dx}_{\text{[2]}}$ $+\displaystyle\underbrace{\ln(2)\int_{0}^{\frac{\pi}{2}}\frac{(-1)^{k}x\cos(2kx)}{k}dx}_{\text{[3]}}$ $+\underbrace{\ln^{2}(2)\int_{0}^{\frac{\pi}{2}}xdx}_{\text{[4]}})dx$

$[1]:\displaystyle \frac{{\pi}^{2}}{16}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{2}}=\frac{-{\pi}^{4}}{192}$

$[2]: \ln(2)\left(\frac{-1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{3}}+\frac{1}{4}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{3}}\right)$

$[3]: \displaystyle \ln(2)\left(\frac{-1}{4}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{3}}+\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{3}}\right)$

$[4]: \displaystyle \ln^{2}(2)\int_{0}^{\frac{\pi}{2}}xdx=\frac{{\pi}^{2}\ln^{2}(2)}{8}$

[2] and [3] cancel one another out and I arrive at:

$\displaystyle\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$

This worked out beautifully. Is it a fluke or can one manipulate sums, like I done above, under certain conditions?. Or did I actually manage to come up with a clever solution?.

Also sorry for the undersized parentheses. Every time I tried enlarging them, the Latex would not display. I have been wrestling with this for sometime trying to get it all to display. Thanks All.

-
You can collapse the sums because of the identities $\int_{-\pi}^{\pi}\cos(kx)\cos(jx)dx=2\int_{0}^{\pi}\cos(kx)\cos(jx)dx=0$ unless $j=k$. This is the basic orthogonality property of Fourier cosine series. – Peder Feb 1 '13 at 21:42
Thanks, Peder. I have since found that out. Thank you. – Cody Feb 2 '13 at 10:12