Using Parseval's to Evaluate an Integral

Question

I have done some exercises on the Parseval's identity and I think it's quite straight forward. However, I came across this exercise and it made me confused. I'll explain:

The function is $f(x) = \pi x - x^2$ on the interval [0, ].

I shall find the sine series of the function and prove that: $$\sum_{n=1}^\infty \frac{1}{n^6} = \frac{\pi^6}{945}$$

Now, I found the correct Fourier coefficients and I get:

$$f(x) = \sum_{n=1}^\infty \frac{4}{\pi n^3}(1-(-1)^n)$$

Perfect. Now, using the Parseval's I want to calculate $|b_n|^2$ and here is where the problem occure. The problem is the $(1-(-1)^n)^2$ term. I tried to evaluate it:

1. $(\frac{4}{\pi n^3}(1-(-1)^n)^2 = \frac{16}{\pi^2 n^6}(1-(-1)^n)^2 = \frac{16}{\pi^2 n^6}(1-2(-1)^n + (-1)^{2n}) = \frac{16}{\pi^2 n^6}(2-2(-1)^n)$ which is 0 for even n!

So what I get is: $$ |b_n|^2 = \begin{cases} \frac{64}{\pi^2(2k+1)^6}, & n \text{ odd} \\\\ 0, & n \text{ even.} \end{cases} $$

Now, how does that help me prove what I should prove???

Answer 1

Since as you wrote

$$1-(-1)^n=\begin{cases}0\,,\,\,n\;\text{is even}\\{}\\ 2\;,\;\;n\;\text{is odd}\end{cases}$$

so that your sine series is

$$f(x)=\sum_{n=1}^\infty\frac{4\cdot2}{\pi(2n-1)^3}\sin nx=\frac8\pi\sum_{n=1}^\infty\frac1{(2n-1)^3}\sin nx$$

And now Parsival:

$$\frac1{2\pi}\int_0^\pi(\pi x-x^2)^2 dx=\frac{64}{\pi^2}\sum_{n=1}^\infty\frac1{(2n-1)^6}\implies \sum_{n=1}^\infty\frac1{(2n-1)^6}=\frac{\pi^6}{15\cdot64}$$

And finally:

$$\sum_{n=1}^\infty\frac1{n^6}=\sum_{n=1}^\infty\frac1{(2n-1)^6}+\sum_{n=1}^\infty\frac1{(2n)^6}=\frac{\pi^6}{15\cdot 64}+\frac1{64}\sum_{n=1}^\infty\frac1{n^6}\implies$$

$$\frac{63}{64}\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{15\cdot 64}\implies\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$$

Check the above so that everything's clear...and correct.