Integral with parameters

Quote:

Originally Posted by DeMath

Calculate

$\int\limits_0^{+\infty}e^{-ax^2} \cos(bx^2)\,dx,~~~a>0$

starting from $\int_0^{\infty} e^{-x^2} \ dx = \frac{\sqrt{\pi}}{2}$ and letting $x^2=st,$ where $s$ is considered to be the variable, we'll get $\frac{1}{\sqrt{t}}=\frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{s}}e^{-st} \ ds. \ \ \ \ \ (1)$

next let $J=\int_0^{\infty} \frac{x+a}{\sqrt{x}[(x+a)^2 + b^2]} \ dx, \ \ a >0.$ let $x=z^2$ to get $J=2\int_0^{\infty} \frac{z^2 + a}{(z^2+a)^2 + b^2} \ dz.$ let $u=z - \frac{\sqrt{a^2+b^2}}{z}, \ v=z+\frac{\sqrt{a^2+b^2}}{z}.$ then:

$2\int \frac{z^2 + a}{(z^2+a)^2 + b^2} \ dz=\left(1+\frac{a}{\sqrt{a^2+b^2}} \right)J_1 + \left (1 - \frac{a}{\sqrt{a^2+b^2}} \right)J_2,$ where $J_1=\int \frac{du}{u^2 + 2\sqrt{a^2+b^2}+2a}$ and $J_2=\int \frac{dv}{v^2 - (2\sqrt{a^2+b^2} - 2a)}.$

both $J_1,J_2$ are easy to find. we'll eventually get: $J=\pi \sqrt{\frac{\sqrt{a^2+b^2} + a}{2(a^2+b^2)}}. \ \ \ \ \ \ \ \ \ (2)$

now let $I=\int_0^{\infty} e^{-ax^2} \cos(bx^2) \ dx, \ \ a>0.$ put $x=\sqrt{t}$ to get $I=\frac{1}{2} \int_0^{\infty} \frac{e^{-at} \cos(bt)}{\sqrt{t}} \ dt = \frac{1}{2\sqrt{\pi}} \int_0^{\infty} e^{-at} \cos(bt) \int_0^{\infty} \frac{1}{\sqrt{s}}e^{-st} \ ds \ dt,$ by (1).

changing the order of integration will give us $I=\frac{1}{2\sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{s}} \int_0^{\infty} e^{-(s+a)t} \cos(bt) \ dt \ ds=\frac{1}{2\sqrt{\pi}} \int_0^{\infty} \frac{s+a}{\sqrt{s}[(s+a)^2 + b^2]} \ ds,$ and therefore by (2):

$I=\sqrt{\frac{\pi(\sqrt{a^2+b^2} + a)}{8(a^2+b^2)}}. \ \ \Box$

the final answer makes sense to me. so i can claim that i have survived from all the annoying algebra involved in there. (Nod)