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The question asks me to prove that $$\int_0^\infty J(x)e^{-ax}dx=\frac{1}{\sqrt{1+a^2}},$$ where $a>0$ and $J(x)=\frac{2}{\pi}\int_0^{\pi/2}\cos(x\cos{\theta})d\theta.$

I started off by considering the function $$f(x,\theta)=\frac{2}{\pi}e^{-ax}\cos(x\cos{\theta})\boldsymbol{1}_{\theta\in(0,\frac{\pi}{2})}$$ in the region $(x,\theta)\in(0,\infty)\times (0,\infty).$ Now $\vert f(x,\theta)\vert\leq \frac{2}{\pi}e^{-ax},$ so $$\int_0^\infty\left(\int_0^\infty\vert f(x,\theta)\vert dx\right)d\theta<\infty,$$ which means that by Fubini's we can do the integration in any order.

I don't think $\cos(x\cos{\theta})$ is integrable in terms of $\theta,$ so I'm thinking I should integrate $e^{-ax}\cos(x\cos{\theta})$ in terms of $x.$ But the integration of this looks quite horrible, so I was wondering (a) if my method is correct, and (b) is there a neater, more efficient way I could solve this problem?

egreg
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quasicoherent_drunk
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  • Beware: $\theta \in (0,\pi/2)$. The context: $J$ is the classical $J_0$ Bessel function and you are asked a classical result concerning the Laplace Transform of this $J_0$. I will probably give an answer later on. – Jean Marie Apr 22 '16 at 12:45
  • Related: http://math.stackexchange.com/questions/326778/laplace-transform-of-the-bessel-function-of-the-first-kind – Marco Cantarini Apr 22 '16 at 12:55
  • @Marco Cantarini : the reference you just gave uses complex function theory (sometimes misnamed "residue calculus"). What dhk628 desires is a solution with classical (lebesgue) integration theory. – Jean Marie Apr 22 '16 at 13:03
  • @MarcoCantarini Thank you for the reference! But as the above comment points out, I am looking for a solution using Lebesgue integration. – quasicoherent_drunk Apr 22 '16 at 13:17
  • @JeanMarie Wouldn't the indicator function in the definition of $f(x,\theta)$ take care of the condition $\theta\in (0,\pi/2)?$ – quasicoherent_drunk Apr 22 '16 at 13:18
  • You are right, I forgot it. – Jean Marie Apr 22 '16 at 13:40

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Let be $$I=\int_0^\infty J(x)e^{-ax}\mathrm dx$$ and $$ J(x)=\frac{2}{\pi}\int_0^{\pi/2}\cos(x\cos{\theta})\mathrm d\theta $$ then we have $$ I=\int_0^\infty \underbrace{\left(\frac{2}{\pi}\int_0^{\pi/2}\cos(x\cos{\theta})\mathrm d\theta\right)}_{J(x)}e^{-ax}\mathrm dx=\frac{2}{\pi}\int_0^{\pi/2}\underbrace{\left(\int_0^{\infty}\cos(x\cos{\theta})e^{-ax}\mathrm d x\right)}_{A(a,\cos\theta)}\mathrm d\theta $$ as you stated.

Observe that $$ A(a,b)=\int_0^\infty e^{-ax}\cos(b x)\mathrm d x=-\frac{e^{-ax}}{a^2+b^2}\left(a\cos(b x)-b\sin(bx)\right)\Big|_0^\infty=\frac{a}{a^2+b^2}\qquad \text{for }a>0 $$ that is $A(a,b)=\mathcal L\{\cos(bx)\}(a)$ is the Laplace transform of $\cos(bx)$.

So for $b=\cos\theta$ we have $$\begin{align} I&=\frac{2}{\pi}\int_0^{\pi/2}A(a,\cos\theta)\mathrm d\theta =\frac{2}{\pi}\int_0^{\pi/2}\frac{a}{a^2+\cos^2\theta}\mathrm d\theta\\ &=\frac{2}{\pi}\left[{\frac{1}{\sqrt{1+a^2}}\tan^{-1}\left(\frac{a \tan x}{\sqrt{1+a^2}}\right)}\right]_0^{\pi/2}=\frac{2}{\pi}\frac{\frac{\pi}{2}}{\sqrt{1+a^2}}=\frac{1}{\sqrt{1+a^2}} \end{align} $$

alexjo
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    $\displaystyle \int e^{-ax} \cos bx , dx = -\dfrac{e^{-ax}}{a^2+b^2}\left(a \cos bx - b \sin bx\right) + C$ I think. –  Apr 22 '16 at 13:07