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It seems to be a pretty simple integral but I am blanking out on how to find this integral:

$\int_0^1 e^{-x}\cos(n\pi x) \;dx$

I tried Mathematica, but it seems to use some formula instead of integrating by parts.

Thanks for your help!

3 Answers 3

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I don't believe it matters which part you integrate and which part you take the derivative of. If you took the derivative of $e^{-x}$ the first time, do it for the second integration by parts as well.

If you have done it correctly, you should have an equation with the same integral on each side, yet they won't cancel out. Simply combine like terms and you should have your answer.

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    ooops, this is not what I actually meant to say. My problem was not seing that after the second integration I had two identical (almost) integrals on the both sides of the equation. I guess it's a good time for me to go to bed. Thanks again!2012-05-27
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When you’ve a product of an exponential and a sine or cosine, you can split it either way. I’ll set $u=e^{-x}$ and $dv=\cos n\pi x\,dx$, so that $du=-e^{-x}dx$ and $v=\frac1{n\pi}\sin n\pi x$. Then

$\begin{align*} \int_0^1 e^{-x}\cos n\pi x\,dx&=\left[\frac1{n\pi}e^{-x}\sin n\pi x\right]_0^1-\int_0^1\frac1{n\pi}\sin n\pi x(-e^{-x})dx\\ &=\frac1{n\pi}\int_0^1 e^{-x}\sin n\pi x\,dx\;. \end{align*}$

Now repeat, making sure to split the new integral the same way: $u=e^{-x}$ and $dv=\sin n\pi x\,dx$, so $du=-e^{-x}dx$, $v=-\frac1{n\pi}\cos n\pi x$, and the integral is

$\begin{align*} \frac1{n\pi}\int_0^1 e^{-x}\sin n\pi x\,dx&=\frac1{n\pi}\left(\left[-\frac1{n\pi}e^{-x}\cos n\pi x\right]_0^1-\frac1{n\pi}\int_0^1e^{-x}\cos n\pi x\,dx\right)\\ &=\frac1{n^2\pi^2}\left(\left[e^{-x}\cos n\pi x\right]_0^1-\int_0^1e^{-x}\cos n\pi x\,dx\right)\;. \end{align*}$

Call the original integral $I$; then

$I=\frac1{n^2\pi^2}\left(\left[e^{-x}\cos n\pi x\right]_1^0-I\right)\;,$

so $\left(1+\frac1{n^2\pi^2}\right)I=\frac1{n^2\pi^2}\left[e^{-x}\cos n\pi x\right]_1^0=\frac1{n^2\pi^2}\left(1-\frac1e\cos n\pi\right)=\frac1{n^2\pi^2}\left(1-\frac{(-1)^n}e\right)\;,$ since $\cos n\pi=(-1)^n$. Finally,

$I=\frac{\dfrac1{n^2\pi^2}\left(1-\dfrac{(-1)^n}e\right)}{1+\dfrac1{n^2\pi^2}}=\frac{e-(-1)^n}{e(n^2\pi^2+1)}\;,$

barring possible careless errors.

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    @dana: You are so right; thanks.2012-05-27
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Here is another method that applies if you know some complex numbers.

Recall

$e^{i \theta} = \cos \theta + i \sin \theta$

In this case we have a $\cos (n \pi x)$, which is the real part of $e^{i n \pi x}$ $\mathscr{R}(e^{in \pi x}) = \cos(n \pi x)$

Then we have $\int_{0}^1 e^{-x} \cos (n \pi x)dx = \mathscr{R} \int_{0}^1 e^{-x}\cdot e^{i n \pi x}$

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    Interesting. Thanks for the tip.2012-05-27