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Find the Fourier coefficients and Fourier series of the following function $$f(x)=\frac{\pi e^{-x}}{e^{\pi}-e^{-\pi}};\quad -\pi\le x\le\pi$$

Here is my work:

Since $L=\pi$ we obtain \begin{align*}a_0&=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(x)\text{d}x=\frac{1}{2\left(e^{\pi}-e^{-\pi}\right)}\int\limits_{-\pi}^{\pi}e^{-x}\text{d}x=\frac{1}{2\left(e^{\pi}-e^{-\pi}\right)}\Big[-e^{-x}\Big]_{-\pi}^{\pi}=\frac{1}{2} \\ a_n&=\frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)\cos\left(nx\right)\text{d}x=\frac{1}{e^{\pi}-e^{-\pi}}\int\limits_{-\pi}^{\pi}e^{-x}\cos(nx)\text{d}x \\ b_n&=\frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)\sin(nx)\text{d}x=\frac{1}{e^{\pi}-e^{-\pi}}\int\limits_{-\pi}^{\pi}e^{-x}\sin(nx)\text{d}x\end{align*}

I'm stuck in $a_{n}$ and $b_{n}$, need help finding an easy way to solve this integration.

Galc127
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user155971
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    What's the problem? You're just integrating sines and cosines, as $e^{-\pi}$ is a constant. – Sean Roberson Aug 22 '17 at 17:15
  • sorry it's e^(-x) – user155971 Aug 22 '17 at 17:16
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    Okay - then it's just integration by parts. Also, as a user with 500+ rep, you should know how to use MathJax and not just paste things in an image. – Sean Roberson Aug 22 '17 at 17:17
  • i'm newbie in this system , sorry , I just need help – user155971 Aug 22 '17 at 17:18
  • Integration by parts with a trick once you arrive at the same integral from integrating by parts twice – JohnColtraneisJC Aug 22 '17 at 17:19
  • Notice that $$f(x)=\underbrace{\frac{\pi}{e^{\pi}-e^{-\pi}}}_{:=m}\cdot e^{-x}$$ and you don't need to carry $m$ in all of your calculations. Are you familiar with integration by parts? Do you know how to evaluate $\displaystyle \int e^{-x}\sin(nx)$? – Galc127 Aug 22 '17 at 17:19
  • i tried with parts but it's very long and didn't get result – user155971 Aug 22 '17 at 17:19
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    Like I said, you need to use a special trick once you arrive at the same integral from integrating by parts twice...the trick is in adding the original integrals together and dividing by 2... – JohnColtraneisJC Aug 22 '17 at 17:21
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    @user155971, I've edited your post. Please notice that you had a mistake in computation of $a_0$. Also, I find that Math Lover's hint is a bit easier than IBP. Up to you choosing what you prefer. – Galc127 Aug 22 '17 at 17:30

2 Answers2

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IBP does the trick. Here's a start:

$$ \frac{1}{n}\int_{-\pi} ^\pi e^{-x} \cos{nx} \ dx = \frac{1}{n} \left(-e^{-x} \cos {nx}\Big|_{-\pi}^\pi - n\int_{-\pi} ^\pi e^{-x} \sin{nx} \ dx \right) $$

The first term is easy. Do another IBP on the second term and you'll get the original integrand, so do what you would do in calculus II and solve for the unknown integral afterwards.

Sean Roberson
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Hint: $$\int e^{-x + i a x} dx = \int{e^{-x} \cos(ax)dx + i \int e^{-x} \sin(ax) dx}.$$

Math Lover
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