Iterated Integrals Question, using Tonelli and/or Fubini

[rainbowed]

Homework Statement

Let J₀(x)=2/$\pi$$\int$₀^$\pi$/2cos(xcos[y])dy. Show that $\int$₀^∞J₀(x)e^-axdx=$\frac{1}{sqrt(1 +a^2)}$.

Homework Equations

Tonelli and Fubini's theorems

The Attempt at a Solution

Basically I'm finding this problem really hard because I've had to teach myself iterated integrals in order to do it, and I'm not sure if I've learned the theory correctly! So far I've tried swapping the integrals (using a combination of Tonelli and Fubini's theorems) which means that I'd have to use integration by parts on e^-ax and cos(xcosy) so that doesn't seem to help... I've also tried not swapping the integrals and using substitution (u=xcosy) but that doesn't make it simpler. Any help would be much appreciated!

 

[mighty2000]

Hi there,

First of all, great job on teaching yourself iterated integrals to tackle this problem! It's definitely not an easy concept to pick up.

To start, let's rewrite J0(x) using the substitution u = cos(y):

J0(x) = 2/π ∫0π/2 cos(xcos[y])dy
= 2/π ∫0π/2 cos(xu)du (since du = -sin(y)dy and when y = 0, u = cos(0) = 1 and when y = π/2, u = cos(π/2) = 0)
= 2/π ∫0π/2 cos(xu)du
= 2/π ∫0π/2 cos(xu)e^-axdx (since we eventually want to integrate with respect to x)

Now, let's use Fubini's theorem to swap the order of integration:

∫0∞J0(x)e^-axdx = ∫0π/2 ∫0∞ cos(xu)e^-axdxdy (since we swapped the order of integration, we are now integrating with respect to x first and then y)

Next, we can use integration by parts to solve this integral. Let u = cos(xu) and dv = e^-axdx. Then du = -usin(xu)dx and v = -1/a e^-ax.

∫0∞J0(x)e^-axdx = ∫0π/2 [uv|0∞ - ∫0∞ vdu]dy
= ∫0π/2 [-1/a cos(xu)e^-ax|0∞ + ∫0∞ (1/a)∫0π/2 sin(xu)du e^-axdx]dy
= ∫0π/2 [-1/a cos(xu)e^-ax|0∞ + ∫0∞ (1/a)∫0π/2 sin(xu)du e^-axdx]dy
= ∫0π/2 [-1/a cos(xu)e^-ax|0∞ + ∫0∞ (1/a) [-cos(xu)|0π/2] e^-axdx]dy
= ∫0π/2 [-1/a cos(xu)e^-ax|0∞