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I found this answer, outlining the exercise, to be interesting. However, I have trouble solving the differential equation.

The question starts by attempting to solve the following integral without complex analysis:

$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$

...So we let

$ F(y) = \int\limits_{0}^{\infty} \frac{\sin xy}{x(1+x^2)} \ dx \ \ \text{for} \quad\quad y > 0$

Next, the portion I'm referring to (from which this question proceeds) starts with

$\displaystyle F''(y) - F(y) + \pi/2 = 0$

I find part of a solution to the differential equation to be

$F(y)=\pi/2+e^y c_1 + e^{-y}c_2$

I'm having trouble finding the constants. Could someone please explain this step in great detail, as I'm somewhat of a novice.

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    @Matt Groff: By bounded I meant that exists a M>0 such that for all $y\geq 0$ we have $|f(y)|\leq M$. I gave the details in an answer.2011-09-11

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As we saw in the comments, we only have to show that $|F(y)|\leq M$ for some $M>0$ independent of $y$. Indeed, if its shown then $|c_1|e^y=|F(y)-\frac{\pi}2-e^{-y}c_2|\leq M+\frac{\pi}2+|c_2|$ for all $y$, therefore $c_1=0$ and $c_2=-\frac{\pi}2$. We have $\displaystyle\int_0^{+\infty}\frac{\sin x}xdx=\int_0^{+\infty}\frac{\sin xy}xdx$ for all $y>0$ (make the substitution $t=xy$). We get \begin{align*} F(y)-\int_0^{+\infty}\frac{\sin x}xdx&=\int_0^{+\infty}\frac{\sin(xy)}x \left(\frac 1{1+x^2}-1\right)dx\\ &=-\int_0^{+\infty}\sin(xy)\frac{x}{1+x^2}dx, \end{align*} and integrating by parts \begin{align*} F(y)-\int_0^{+\infty}\frac{\sin x}xdx&=-\left[\frac 1y\cos(xy)\frac x{1+x^2}\right]_{x=0}^{x\to+\infty}+\int_0^{+\infty}\frac 1y\cos(xy)\left( \frac 1{1+x^2}-\frac{2x^2}{(1+x^2)^2}\right)dx\\ &=\frac 1y\int_0^{+\infty}\frac{\cos(xy)}{1+x^2}dx-\frac 1y\int_0^{+\infty} \cos(xy)\frac{2x^2}{(1+x^2)^2}dx. \end{align*} Finally $ \left|F(y)-\int_0^{+\infty}\frac{\sin x}xdx\right|\leq \frac 1y\frac{3\pi}2,$ hence $F$ has a limit at $+\infty$ (and is continuous): it's bounded.