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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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You have to use the fact that $F(0)=0$ and $F$ is bounded: you will get $c_1=0$ and $c_2=-\frac{\pi}2$. –  Davide Giraudo Sep 10 '11 at 8:07
    
I think you need to copy more of the original detail into this question to make it self-contained - it contains no information other than the DE, and you need some boundary conditions to identify the constants. What might those be? Well one thing to do is to try a reality check and see what you expect to happen to the solution as $y$ gets very large (boundary at infinity). And then can you think about whether it would be legitimate to set $y=0$ (in spite of $y>0$ being specified. –  Mark Bennet Sep 10 '11 at 8:19
    
@Davide Giraudo: I think I see how to get $F(0)=0$, but I don't understand boundedness. What is "bounded", and how do I know that $F$ is bounded? –  Matt Groff Sep 10 '11 at 8:27
1  
Into the integral expression, see what happens as $y\to\infty$, you should observe that it's values are bounded in magnitude, because regardless of what $x,y$ are, $| \sin xy|\leq 1$. So, F(y) is bounded, which places a restriction on what the coefficient of $e^y$ can be. –  Ragib Zaman Sep 10 '11 at 9:00
    
@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. –  Davide Giraudo Sep 11 '11 at 16:20

1 Answer 1

up vote 2 down vote accepted

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.

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