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An exercise I'm trying to solve gives us

$$F(y) := \int_{0}^{\infty} \frac{sin(xy)}{x(1+x^2)}dx$$ and first asks us to prove that F(y) satisfies the ODE $F''(y)-F(y)+\frac{\pi}{2}=0$, then find the closed form of the ODE and lastly prove that:

$$\int_{0}^{\infty}\frac{cos(yx)}{1+x^2}dx=\frac{\pi e^{-y}}{2}$$

using the closed form of the ODE.

I've gone through the first two and for the closed form I got $F(y)=c_1e^{-y}+c_2e^y+\frac{\pi}{2}$.

But I'm having trouble with solving the last part of the exercise. With complex analysis, I can solve it, but I don't know what to do with the closed form. I'm sharing my original thoughts, but I get stuck:

First, I noticed that the last integral is the derivative of F(y):

$$F'(y)= \int_{0}^{\infty} \frac{d}{dy} \Big(\frac{sin(xy)}{x(1+x^2)} \Big)dx= \int_{0}^{\infty}\frac{cos(yx)}{1+x^2}dx $$

Also: $F'(y)=-c_1e^{-y}+c_2e^y$. Then:

$$ \int_{0}^{\infty}\frac{cos(yx)}{1+x^2}dx = -c_1e^{-y}+c_2e^y $$

Then I'd need to show $ -c_1e^{-y}+c_2e^y = \frac{\pi e^{-y}}{2}$ and I get stuck here.

Am I doing something wrong? Is this even solvable this way? Should I approach it differently?

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    $\begingroup$ So you just need to fix c1 and c2? Just try to evaluate the boundary terms F(0) and F'(0). This will fix the particular solution to the ODE that solves your integral.... $\endgroup$
    – Thomas
    Dec 12, 2021 at 19:10

1 Answer 1

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If you proven the first parts already, then all you need is a limiting condition.

$$F'(0) = \int_0^\infty \frac{\cos(0)}{(1+x^2)}dx = \frac{\pi}{2}$$

and consider the change of variables $z=xy$

$$F(y) = \int_0^\infty\frac{\sin z}{z\left(1+\frac{z^2}{y^2}\right)}dz \implies \lim_{y\to\infty}F(y) = \int_0^\infty\frac{\sin z}{z}\:dz = \frac{\pi}{2}$$

The exponentially increasing term must be $0$ since the limit exists at all, and then we have

$$F'(y) = \frac{\pi e^{-y}}{2}$$

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  • $\begingroup$ Oh!! It makes sense now! It didn't even cross my mind to take a condition! Thank you~ $\endgroup$
    – Tita
    Dec 12, 2021 at 21:21

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