I'm sure easier proofs exist, but I have to specifically use the method in the picture:
This is what my attempt is:
First, I did some manipulation to figure out that $$ \sum_{k=1}^\infty \frac 1{k^4} = \frac {16}{15} \sum_{j=1}^\infty \frac 1{(2j-1)^4}$$
The latter sum is more similar to the summations in the problem, so I felt like this was the right thing to start with, and would come in handy toward the end of the problem. Next, I used the result of part (a) to say
$$ \int_{-\pi}^{\pi} \left| \frac {\pi}{2} - \frac {4}{\pi} \sum_{j=1}^{\infty} \frac {Cos((2j-1)x)}{(2j-1)^2} \right| ^2 dx = \int_{-\pi}^{\pi} \lvert x \rvert ^2 dx = \frac {2 \pi^3}{3} $$
Next, I figured that the absolute value sign on the left seemed redundant since we are squaring anyway, so I dropped the absolutes. I wanted to move the integral inside, but the squared sign was stopping me. The only way I could figure to deal with that was to factor everything out. I got:
$$ \int_{-\pi}^{\pi} \left( \frac {\pi^2}{4} - 4 \sum_{j=1}^{\infty} \frac {Cos((2j-1)x)}{(2j-1)^2} + \frac {16}{\pi^2} \left( \sum_{j=1}^{\infty} \frac {Cos((2j-1)x)}{(2j-1)^2}\right)^2 \right)dx = \frac {2 \pi^3}{3} $$
I then distributed the integral. I found the value of the first term and moved it over to the other side. I also found that the middle term (first summation) vanishes. So all I was left with was:
$$ \int_{-\pi}^{\pi} \left(\sum_{j=1}^{\infty} \frac {Cos((2j-1)x)}{(2j-1)^2} \right)^2 dx = \frac {\pi^5}{96} $$
And at this point I got stuck because, again, I couldn't move the integral past the squared sign... and I'm sure I messed up somewhere along the way anyway.... Ideas? :3
