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Consider the $2\pi$-periodic odd function defined on $[0,\pi]$ by $f(\theta)=\theta (\pi - \theta)$.

(b)Compute the Fourier Coefficients of $f$, show that$$f(\theta) = \frac{8}{\pi}\sum_{k\ odd\ \ge 1}\frac{\sin(k\theta)}{k^3} $$

I calculated $\hat{f}(0)$ and it was $\frac{\pi^2}{12}$(which at the end I knew that it is wrong).But the calculation of $\hat{f}(n)$ gave me $$(\frac{1}{2} + \frac{2}{in}) \{\frac{\pi}{-in} e^{-in\pi} + \frac{1}{n^2}(e^{-in\pi} - 1)\} + \frac{1}{2in\pi} (\pi^2 e^{-in\pi})$$(which was also wrong ), which I do not know how it will reach the formula that he wants? and also I do not know the importance of mentioning that the function is odd. May be the book want us to use the following problem results, exactly letter (b) of it Basic Properties Of Fourier Series. 2.. For the integration by parts that we do when we are calculating the Fourier Coefficient, The following link contains the details Integration By Parts Example., Now I knew the correct answer of this question, thanks to all who helped me and I hope this question will be a reference to questions that appear in minds before trying to solve questions like it. The newest and the most beautiful thing that I have learned from this question is adjusting the definition of the given function.

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    It seems you misunderstood the function. $f$ is supposed to be odd and $2\pi$-periodic, and on the interval $[0,\pi]$ is given by $f(\theta) = \theta(\pi - \theta)$. From the oddness, it follows that on the interval $[-\pi,0]$ it is given by $f(\theta) = \theta(\pi + \theta)$. In a single formula, $f(\theta) = \theta(\pi - \lvert\theta\rvert)$ for $\theta \in [-\pi,\pi]$, and then extend it using the periodicity. – 2017-02-28
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    Thank u so much your answer seems so reasonable. but am I going to calculate Fourier Series on$ [0,\pi]$or on $[-\pi, \pi]$. – 2017-02-28
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    Also, I think on the interval $[-\pi, 0]$ $f(\theta) = -\theta (\pi - \theta)$ not $f(\theta) = \theta (\pi + \theta)$ as u wrote, because $f(\theta)$ is negative in this case. – 2017-02-28
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    How can I extend the summation of Fourier Series on$$ k\ odd\ \ge 1$$ instead of $$[-\infty , \infty]$$. – 2017-02-28
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    No .$f(\theta)=\theta(\pi + \theta)$ on $[-\pi , 0]$ as written before by @DanielFischer – 2017-03-01

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First we make f a $2-\pi$ periodic odd function (Extend $f$).

From oddness $f(-\theta)=-f(\theta)$

So on $[-\pi , 0]$ we have $$f(-(-\theta))=-(f(-\theta))=-(-\theta(\pi + \theta))=\theta(\pi + \theta)$$

$f(\theta)=\theta(\pi - |\theta|)$ on $[-\pi , \pi]$

So $$ f(\theta)=\left\{\begin{matrix} \theta(\pi + \theta) \ \ \ \ \theta \in [-\pi ,0]& \\ \theta(\pi - \theta) \ \ \ \ \theta \in [\ \ 0 ,\pi\ ]& \end{matrix}\right. $$

Which implies that : $$ f'(\theta)=\left\{\begin{matrix} \pi + 2\theta \ \ \ \ \theta \in [-\pi ,0]& \\ \pi - 2\theta \ \ \ \ \theta \in [\ \ 0 ,\pi\ ]& \end{matrix}\right. $$

To Calculate the Fouries coefficient :

$$\hat{f}(n)=\frac{1}{2\pi} \int_{-\pi}^{\pi}f(\theta)e^{-in\theta} \ d\theta$$ Integrate by parts implies : $$\hat{f}(n)=\frac{1}{2\pi}\left[\frac{-f(\theta)}{in}e^{-in\theta}\right]_{-\pi}^{\pi} \ + \ \frac{1}{2ni\pi} \int_{-\pi}^{\pi}f'(\theta)e^{-in\theta} \ d\theta=\frac{1}{2in\pi}\left( \int_{-\pi}^{0}(\pi + 2\theta)e^{-in\theta} \ d\theta + \int_{0}^{\pi}(\pi - 2\theta)e^{-in\theta}\right)$$

Which implies that : $$\hat{f}(n)=\left\{\begin{matrix} \frac{4}{n^3\pi i} & n \ is \ odd \\ 0 & n \ is\ even \end{matrix}\right.$$

Finaly : $$f\sim \sum_{n \ odd \ \ge 1}^{} \frac{4}{n^3 \pi i} e^{in\theta}$$

Then From : $$f(\theta) \thicksim \hat{f}(0) + \sum_{n\ge1}[\hat{f}(n) + \hat{f}(-n)]\ cos n\theta + i[\hat{f}(n) - \hat{f}(-n)]\ sin n\theta. $$

You will get : $$f(\theta) = \frac{8}{\pi}\sum_{n\ odd\ \ge 1}\frac{\sin(n\theta)}{n^3} $$

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    Then by substituting with the value of $\hat{f}(n)$ in the third line from below in the answer we will get the required. Thanks to all who helped me. – 2017-03-03
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    You're welcome . – 2017-03-04