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This integral resulted from trying to solve a physics problem about diffusion: $$\int^L_{-L}\sum_{n=1}^{\infty} \cos\Big( \frac{(2n-1)\pi}{L}x\Big)e^{-D\left(\frac{(2n-1)\pi}{L}\right)^2t}\;dx $$ I thought about interchanging the integral sign and the summation and then just having $$\sum_{n=1}^{\infty}e^{-D\left(\frac{(2n-1)\pi}{L}\right)^2t}\int^L_{-L} \cos\Big( \frac{(2n-1)\pi}{L}x\Big)\;dx $$ Since $$\int^L_{-L} \cos\Big( \frac{(2n-1)\pi}{L}x\Big)\;dx = \frac{2 L \sin (2 π n)}{π-2 π n}$$ then my original integral is just equal $0$. I don't know if interchanging the integral with the summation is allowed in this case but I've seen it done before.

EDIT: Here is the original problem:

$$ \frac{dC}{dt} = D\frac{d^2C}{dx^2} $$ with $$ C(L,t) = C(-L,t) \\ \frac{dC}{dx}(L,t) = \frac{dC}{dx}(-L,t) = 0 \\ C(x,0) = \delta(x) $$ The integral is not the final solution but rather part of it.

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  • $\begingroup$ Note that $\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx$. The integral is linear and can be "dispersed" into a summation. $\endgroup$ – Kaynex Nov 9 '16 at 3:27
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    $\begingroup$ Could you tell us how you arrived at the above integral? $\endgroup$ – Jacky Chong Nov 9 '16 at 3:29
  • $\begingroup$ You might want to read: math.stackexchange.com/q/188567/9464 $\endgroup$ – Jack Nov 9 '16 at 3:34
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    $\begingroup$ @JackyChong It results from solving the heat equation in one dimension with homogeneous boundary conditions and an initial condition equal to the Dirac-Delta function. $\endgroup$ – Lucas Alanis Nov 9 '16 at 3:34
  • $\begingroup$ Neat source, Jack. I didn't know something like this was so complicated. $\endgroup$ – Kaynex Nov 9 '16 at 3:37
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The exponential part is irrelevant and the constant function $1$ is continuous on the interval $[-L,L]$. Hence, you can integrate its Fourier series term by term.

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