# Laplace transform and integration together

The question, given in the textbook, is somewhat different. However, I am rephrasing it as follows:

$$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}} \left[ \mathcal L \lbrace \cos(t\cos\theta) \rbrace \right]dt$$

The book then uses the standard table of Laplace transform to arrive at the below expression:
$$\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}} \frac{s}{s^2 + \cos^2\theta} d\theta$$

Then, I do not understand what to do. The book has it all messed up.
Can someone please tell me, step-by-step, what to do next ?

The original question in the book is: Let

$$f(t) = \frac{1}{\pi} \int_{0}^{\pi}{\cos(t\cos\theta)~ \mathrm d\theta}.$$ Prove that: $$\mathcal L[f(t)] = \frac{1}{\sqrt{s^2+1}}.$$

• Shouldn't the integral be in terms of $s$? – yousuf soliman Aug 24 '13 at 15:32
• @Pixelwyrm Please see the edit. I posted the original question in the book. – Little Child Aug 24 '13 at 15:36

You want to evaluate

$$\frac{2}{\pi} s \int_0^{\pi/2} \frac{d\theta}{s^2+\cos^2{\theta}}$$

This may be done through conversion to a contour integral in the complex plane from which we may apply the residue theorem. First note that we may extend the integration interval to $[0,2 \pi]$ by symmetry. Let $z=e^{i \theta}$, then $d\theta = -i dz/z$ and $\cos{\theta} = (z+1/z)/2$. Therefore we have the integral being equal to

$$\frac{s}{i 2 \pi} \oint_{|z|=1} \frac{dz}{z} \frac{1}{s^2+\frac14 (z+z^{-1})^2} = \frac{2 s}{i \pi} \oint_{|z|=1} dz \frac{z}{z^4+(4 s^2+2) z^2+1}$$

The integrand has 4 poles, at $z=\pm i \left ( \sqrt{s^2+1}+s\right)$ and $z=\pm i/\left(\sqrt{s^2+1}+s\right)$. Note that only the latter two poles are in the interior of the unit circle $|z|=1$, so we need only compute residues at those poles. Noting that, for a function of the form $p(z)/q(z)$ with simple pole at $z=z_0$, the residue at that pole is $p(z_0)/q'(z_0)$, we find that the integral is, by the residue theorem, the sum of the residues at the poles inside the unit circle, or

$$i 2 \pi \frac{2 s}{i \pi} \frac{2}{-4 (\sqrt{s^2+1}-s)^2+4 (2 s^2+1)} = \frac{1}{\sqrt{s^2+1}}$$

• The book simply multiplied the top by $Sec^2\theta$ and proceeded on with normal integration. Yours is way different from the book but provides a useful alternative solution :) – Little Child Aug 24 '13 at 17:20

The first thing we can note is that $f(t)$ is the Bessel function $J_0(x)$. It is necessary to know that the Bessel function of order $0$ is defined as the solution to: $$x^2y''+xy'+x^2y=0$$ I decided to go about solving the original problem slightly differently.

The Laplace transform of the equation $x(y''+y)+y'=0$ gives us: \begin{align}0&=-\frac{d}{ds}\mathcal L\{y''+y\}+\mathcal L\{y'\}\\&=-\frac{d}{ds}(s^2\mathcal L\{y\}(s)+\mathcal L\{y\}(s)-sy(0)-y'(0))+s\mathcal L\{y\}(s)-y(0)\\&=-(1+s^2)\mathcal L'\{y\}(s)-s\mathcal L\{y\}(s)\end{align}

We now have $$\mathcal L\{J_0\}(s)=\frac{c}{1+s^2}$$ We can compute $c$ through $$0=\lim_{x\rightarrow\infty}\mathcal L\{J^\prime_0\}(x)=c-1\implies c=1$$

Therefore $$\mathcal L\{f(t)\}=\frac{1}{\sqrt {1+s^2}}$$