Need help with this calculus problem.
Compute the integral: $\int^{\infty}_{-\infty}(p+x^2)^{-3} dx$ where $p$ is a fixed positive number.
May someone give full solution.It is in here (HW#1) and it is not homework because it is due a month ago.
Need help with this calculus problem.
Compute the integral: $\int^{\infty}_{-\infty}(p+x^2)^{-3} dx$ where $p$ is a fixed positive number.
May someone give full solution.It is in here (HW#1) and it is not homework because it is due a month ago.
As usual, I try a special function approach to the problem.
You have: $\tag{1} \begin{split} \int_{-\infty}^\infty (p+x^2)^{-3}\ \text{d} x &= \frac{2}{p^3}\ \int_0^\infty (1+x^2/p)^{-3}\ \text{d} x \\ &\stackrel{u=x^2/p}{=} \frac{1}{p^{5/2}}\ \int_0^\infty \frac{u^{-1/2}}{(1+u)^3}\ \text{d} u\; ;\end{split}$ keeping in mind that the integral defining the beta function $B(t,s)$ can be put in the form: $B(t,s)=\int_0^\infty \frac{u^{t-1}}{(1+u)^{t+s}}\ \text{d} u$ when $t,s>0$, the rightmost side of (1) equals $1/p^{5/2}\ B(1/2,5/2)$. Finally, since: $B(t,s)=\frac{\Gamma (t)\ \Gamma (s)}{\Gamma (t+s)}\; ,$ $\Gamma (1/2)=\sqrt{\pi}$ and: $\Gamma (1+t)=t\ \Gamma (t)\qquad \text{(hence } \Gamma (1+n)=n! \text{ for positive integers)}$ (where $\Gamma (\cdot)$ is Euler gamma function), you get: $\begin{split} \int_{-\infty}^\infty (p+x^2)^{-3}\ \text{d} x &= \frac{1}{p^{5/2}}\ \frac{\Gamma (1/2)\ \Gamma (5/2)}{\Gamma (3)}\\ &= \frac{1}{p^{5/2}}\ \frac{\sqrt{\pi}\ (\frac{3}{2}\ \frac{1}{2}\ \sqrt{\pi})}{2!}\\ &= \frac{3\pi}{8\ p^{5/2}}\; .\end{split}$
Hint: Let $x=\sqrt{p}\tan\theta$. Then $dx=\sqrt{p}\,\sec^2\theta\,d\theta$. Our integral after a while becomes $\int_{-\pi/2}^{\pi/2} \frac{\sqrt{p}}{p^3} \cos^4 \theta\,d\theta.$ The integration is now maybe familiar. One can do it by parts, or by using double-angle identities.
Hint: Differentiate twice with respect to $a$ both sides of the identity $\int^{\infty}_{-\infty} \frac{1}{x^2+a^2} dx = \frac{\pi}{a} .$