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There's a problem (#10.23) in Apostol's Mathematical Analysis of which I am having a rough time solving: Let $F(y)= \int_{0}^{\infty}\frac{\sin xy}{x(x^{2}+1)}dx$ if $y > 0$. Show that $F$ satisfies the differential equation $F''(y)-F(y)+\frac{\pi }{2} = 0$ and deduce that $F(y)= \frac{1}{2}\pi(1-e^{-y})$. Use this result to deduce the following equations, valid for $y > 0$ and $a > 0$: Use this to deduce that for $y>0$ and $a>0$ \begin{align} \int_{0}^{\infty}\frac{\sin xy}{x(x^{2}+a^{2})}dx = \frac{\pi}{2a^{2}}(1-e^{-ay}), \newline \int_{0}^{\infty}\frac{\cos xy}{x^{2}+a^{2}}dx = \frac{\pi e^{-ay}}{2a}, \newline \int_{0}^{\infty}\frac{x\sin xy}{x^{2}+a^{2}}dx = \frac{\pi}{2}e^{-ay} \end{align} We could use $\int_{0}^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$

How can I show this?

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You can use dollar signs to get your tex code compiled. –  Rasmus Apr 20 '11 at 21:39
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this problem has multiple parts. Which one are you stuck at? –  Alon Amit Apr 20 '11 at 21:43
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Which parts do you have trouble with? –  Rasmus Apr 20 '11 at 21:43
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This is posted as an answer to a previous question: math.stackexchange.com/questions/9402/… . –  Raeder Apr 20 '11 at 21:57
    
@user9848: Assume all the nice things like swapping the derivative with the integral etc. Are you having trouble with justifying this or with the algebra part? –  user17762 Apr 20 '11 at 22:35

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Let $f(x,y)=\frac{\sin(xy)}{x(x^2+1)}$. Then $$f_{yy}(x,y)-f(x,y)=-\frac{x\sin(xy)}{x^2+1}-\frac{\sin(xy)}{x(x^2+1)}=-\frac{\sin(xy)}{x}$$ It follows that $$F''(y)-F(y)=\int_0^\infty -\frac{\sin(xy)}{y} dx=-\int_0^\infty \frac{\sin(xy)}{xy} \,d(xy)=-\pi/2$$ Then $F(y)$ is a solution to the differential equation $z''-z+\frac{pi}{2}=0$. It follos that $F(y)=ae^y+be^{-y}+\frac{\pi}{2}$. From the definition of $F(y)$ one can check that $\lim_{y\to 0} F(y)=0$ and $\lim_{y\to 0} F'(y)=\pi/2$. Then $a+b+\pi/2=0$ and $a-b=\pi/2$. Solving for $a$ and $b$ we have $a=0$ and $b=-\pi/2$. Therefore $$F(y)=\frac{\pi}{2}(1-e^{-y})$$. Let $$G(y):=\int_{0}^{\infty}\frac{\sin xy}{x(x^{2}+a^{2})}dx.$$By substitution $t=x/a$ one can show that $G(y)=\frac{F(ay)}{a^2}=\frac{\pi}{2a^2}(1-e^{-ay})$ and then $$\begin{align*}G'(y)&=\frac{F'(ay)}{a}=\frac{\pi e^{-ay}}{2a}\\G''(y)&=F''(ay)=-\frac{\pi}{2}e^{-ay}\end{align*}$$

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