Compute the value of the integral $\int ^{\infty}_0 \frac{1}{1+2ax+x^2} dx $

Question

Compute the value of the integral $$\int ^{\infty}_0 \frac{1}{1+2ax+x^2} dx $$ Differentiating by $a$ didn't work for me, and I'm all out of ideas..

Answer 1

If you complete the square then the integral becomes, $$\int ^{\infty}_0 \frac{1}{1+2ax+x^2} dx=\int_0^\infty\frac{1}{(x+a)^2+1-a^2}$$

For $a=1$ the solution is obvious.

For $a \gt 1$, this will lead to $$\int_0^\infty\frac{1}{(x+a)^2-(\sqrt{a^2-1})^2}$$ which can be evaluated.

For $a \lt 1$, the integral will turn out to be $$\int_0^\infty\frac{1}{(x+a)^2+(\sqrt{1-a^2})^2}$$ which also can be evaluated.

Answer 2

Assume for definiteness that $0\leq a<1$. Then, after the change of variables $x=-a+\sqrt{1-a^2}\tan t$ the integral transforms into $$\frac{1}{\sqrt{1-a^2}}\int_{\arctan\frac{a}{\sqrt{1-a^2}}}^{\pi/2} dt= \frac{\arccos a}{\sqrt{1-a^2}}.$$ For $a>1$, the appropriate change of variables would be $x=-a+\sqrt{a^2-1}\coth t$, with the final result given by $\displaystyle \frac{\operatorname{arccosh} a}{\sqrt{a^2-1}}$.