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Let $c>0$. Use the Plancherel formula to compute the integral $$\int_{-\infty}^\infty \dfrac{1}{c^2+y^2}dy$$

We note that the Fourier transform of the function $f(x)=e^{-cx}$ for $x\geq 0$ and $0$ for $x<0$ is $\dfrac{1}{c+iy}$. (This follows from direct computation.)

The Plancherel formula says that $\|\hat{f}\|_2^2=2\pi\|f\|_2^2$. Writing out fully, this means $$\int_{-\infty}^\infty \hat{f}(y)^2dy=2\pi\int_0^\infty f(x)^2dx$$ Plugging in $f$ and $\hat{f}$, this yields $$\int_{-\infty}^\infty \dfrac{1}{(c+iy)^2}dy=2\pi\int_0^\infty e^{-2cx}dx$$ I don't see how this helps with the original integral.

Edit.. thanks to Daniel Fischer's comment, it should be $$\int_{-\infty}^\infty \dfrac{1}{|c+iy|^2}dy=2\pi\int_0^\infty e^{-2cx}dx$$

The integral on the right is $\dfrac{1}{-2c}e^{-2cx}|_0^\infty = \dfrac{\pi}{c}$

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    $\begingroup$ It's not $(c+iy)^2$ in the denominator but $\lvert c+iy\rvert^2$. (Generally, $\int \lvert \hat{f}(y)\rvert^2\,dy$ and $\int \lvert f(x)\rvert^2\,dx$.) $\endgroup$ Nov 9, 2013 at 20:54
  • $\begingroup$ @DanielFischer Wow, that's very important! $\endgroup$
    – PJ Miller
    Nov 9, 2013 at 20:55

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It's not just the square of $\hat{f}$, since $\hat{f}$ takes on complex values; you have to multiply by the complex conjugate: Recall that the $L^2$ inner product is given by $$(f, g) = \int f \overline{g}$$

So in this case, the complex conjugate is $1 / (c - iy)$, so we actually get

$$\int_0^{\infty} e^{-2cx} = \int_{-\infty}^{\infty} \frac{1}{(c - iy)(c + iy)} = \int_{-\infty}^{\infty} \frac{1}{c^2 + y^2}$$

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