Integral over the real axis using complex analysis

Question

The problem at hand is to compute the integral $$ \int_{-\infty}^{\infty}dx \dfrac{1}{\sqrt{(x+a)^2+b^2}\sqrt{(x-a)^2+b^2}} $$ where $a$ and $b$ are both real numbers. The square roots introduce branch cuts in the complex plan, so we can't simply use the residue theorem. How can I evaluate this integral? Thanks!

Answer 1

If for any $a\geq 0$ and $b>0$ we define $$ F(a,b)=\int_\mathbb{R}\frac{dx}{\sqrt{(x-a)^2+b^2}\sqrt{(x+a)^2+b^2}} $$ we have $F(0,b)=\frac{\pi}{b}$ and $$ F(a,b) = a\int_\mathbb{R}\frac{dx}{\sqrt{a^2(x-1)^2+b^2}\sqrt{a^2(x+1)^2+b^2}}=F\left(1,\tfrac{b}{a}\right) $$ for any $a>0$. In such a case, by letting $\lambda=\frac{b}{a}$ we have $$ F(1,\lambda)=\int_{\mathbb{R}}\frac{dx}{\sqrt{(x^2-1)^2+2\lambda^2(x^2+1)+\lambda^4}}=\int_{\mathbb{R}}\frac{dx}{\sqrt{(x^2+\lambda^2-1)^2+4\lambda^2}} $$ or $$ F(1,\lambda)=\int_{\mathbb{R}}\frac{dx}{\sqrt{[x^2+(\lambda+1)^2]\cdot[x^2+(\lambda-i)^2]}}=\frac{\pi}{2\,\text{AGM}(\lambda+i,\lambda-i)} $$ or (assuming $\lambda>1$) $$ F(1,\lambda)=\frac{\pi}{2\,\text{AGM}(\lambda,\sqrt{\lambda^2-1})}=\frac{1}{\lambda}\,K\left(\tfrac{1}{\lambda^2}\right) $$ where $K$ is the complete elliptic integral of the first kind and $\frac{1}{\lambda^2}$ is the elliptic modulus.
It follows that $b>a>0$ leads to $$ F(a,b) = \frac{a}{b}\,K\left(\frac{a^2}{b^2}\right).$$