Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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 $b_n$ coefficients, 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.

share|cite|improve this question
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
up vote 4 down vote accepted

When you integrate (using integration by parts or a table) \begin{equation*} b_n=\int_{-1}^{1}x^3\sin (n \pi x) \, dx = 2\int_{0}^{1}x^3 \sin (n \pi x) \, dx \end{equation*} you get \begin{equation*} -\frac{2(-1)^n}{n\pi}\left( 1-\frac{6}{n^2\pi^2} \right). \end{equation*} Squaring this, we obtain \begin{equation*} \frac{4}{n^2\pi^2}-\frac{48}{n^4\pi^4}+\frac{144}{n^6\pi^6}. \end{equation*} If we use the fact that \begin{equation*} \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}, \quad \sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90} \end{equation*} together with Parseval's identity \begin{equation*} \frac{1}{1}\int_{-1}^{1}|x^3|^2 \, dx = \sum_{n=1}^{\infty}|b_n|^2\quad (\text{since } a_n=0) \end{equation*} the result follows.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.