# Using Parseval' s theorem to evaluate a sum..

On the function $f(x)$ = $x^3$ on $(-1,1)$ find Fourier coefficients for this function and then use Parseval's Theorem to evaluate:

$$\sum_{n=1}^{\infty}\frac{1}{(n^6)}.$$

Current work:

I have used Mathematica to find the sum, which results in $\frac{\pi^6}{945}$

Since $a_0$ and $a_n = 0$ as it is an odd function, my problem is calculating $b_n$. I am using Mathematica and resulting in:

$$\int_a^b x^3 \sin{(n x \pi)} \,dx= \frac17$$ in which a is -1 and b is 1, the right hand side is calculated from the average value formula, by squaring Integral$[f(x)dx]$.

This is the problem, as my result for the $b_n$ coeff, is resulting in a nasty rational:

$$-2(-6+n^2\pi^2)cos(n\pi) / n^3\pi^3$$

where I can't seem to factor out a $\frac{1}{n^6}$ term to evaluate the sum using the theorem.

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are you sure $\int_{-1}^1 x^3 \sin(nx\pi)dx = \frac{1}{7}$, it contradicts Riemann-Lesbegue Lemma. –  Yimin Feb 15 '13 at 19:36
sorry, what i meant from that was that, there are two sides of the theorem's equation. the left hand side which computes $b_n$ which is equal to the average value of just $f(x)^2$, which computes to 1/7. The left hand side of the equation, the integral, is just set up in the equation form, unless I am doing something wrong.. –  julesverne Feb 15 '13 at 19:41

superb.. thank you so much. I was stuck after the term expansion, not knowing that using the $1/n^2$ identity to help reduce would essentially cancel out all the unnecessary terms. –  julesverne Feb 15 '13 at 21:09