Integral of $\int_{-\infty}^{\infty} \left(\frac{1}{\alpha + ix} + \frac{1}{\alpha - ix}\right)^2 \, dx$

Question

I'm having trouble integrating

$$\int_{-\infty}^{\infty} \left(\frac{1}{\alpha + ix} + \frac{1}{\alpha - ix}\right)^2 \, dx$$

where $\alpha$ is a real number and $i = \sqrt{-1}$. I'm guessing that I would have to integrate it using methods on complex analysis wherein I would use a semicircle contour enclosing a pole ($x = i\alpha$ for example) so that I could apply the residue theorem. But is there any other way to evaluate this integral (like an ingenious substitution or differentiating under the integral sign)?

Answer 1

Observe that the second summand is conjugate of the first: $$ \frac1{\alpha-ix}=\overline{\frac1{\alpha+ix}} $$ thus their sum will be real. In particular $$ \frac1{\alpha+ix}+\frac1{\alpha-ix}=\frac{2\alpha}{\alpha^2+x^2}{} $$ thus the integrand will be $$ \frac{4\alpha^2}{(\alpha^2+x^2)^2}=\frac{4}{\alpha^2}\frac{1}{(1+(x/\alpha)^2)^2} $$ The substitution $x=\alpha\tan y$ leads to (suppose wlog $\alpha>0$) \begin{align*} \int_{-\infty}^{+\infty}\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\,dx &=\frac{4}{\alpha^2}\int_{-\infty}^{+\infty}\frac{dx}{(1+(x/\alpha)^2)^2}\\ &=\frac{4}{\alpha^2}\int_{-\pi/2}^{+\pi/2}\frac{1}{(1+\tan^2y)^2}\frac{\alpha}{\cos^2y}\,dy\\ &=\frac{4}{\alpha}\int_{-\pi/2}^{+\pi/2}\frac{1}{\frac{1}{\cos^4y}}\frac{1}{\cos^2y}\,dy\\ &=\frac{4}{\alpha}\int_{-\pi/2}^{+\pi/2}{\cos^2y}\,dy\\ &=\frac{4}{\alpha}\frac12\left[\cos y\sin y +y\right]_{-\pi/2}^{+\pi/2}\\ &=\frac{2\pi}{\alpha} \end{align*}

Answer 2

The integral is even in $\alpha$, so assume that $\alpha\gt0$.

Real Analysis Approach

Substitute $x\mapsto\alpha x$, $$ \begin{align} \int_{-\infty}^\infty\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\mathrm{d}x &=\int_{-\infty}^\infty\left(\frac{2\alpha}{\alpha^2+x^2}\right)^2\mathrm{d}x\\ &=\frac4\alpha\int_{-\infty}^\infty\frac1{(1+x^2)^2}\mathrm{d}x\tag{1} \end{align} $$ Furthermore $$ \begin{align} \pi &=\int_{-\infty}^\infty\frac1{1+x^2}\,\mathrm{d}x\tag{2}\\ &=\int_{-\infty}^\infty\frac{2x^2}{(1+x^2)^2}\,\mathrm{d}x\tag{3}\\ &=\int_{-\infty}^\infty\frac{1+x^2}{(1+x^2)^2}\,\mathrm{d}x\tag{4}\\ &=\int_{-\infty}^\infty\frac2{(1+x^2)^2}\,\mathrm{d}x\tag{5}\\ \end{align} $$ Explanation:
$(2)$: arctan integral
$(3)$: integration by parts on $(2)$
$(4)$: multiply the integrand in $(2)$ by $\frac{1+x^2}{1+x^2}$
$(5)$: $2$ times $(4)$ minus $(3)$

Plug $(5)$ into $(1)$ to get $$ \begin{align} \int_{-\infty}^\infty\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\mathrm{d}x &=\frac4\alpha\frac\pi2\\[3pt] &=\frac{2\pi}{\alpha}\tag{6} \end{align} $$

---

Complex Analysis Approach

Let $z=ix$ and $\gamma=[-iR,iR]\cup iRe^{i[0,\pi]}$ as $R\to\infty$.

Then $$ \begin{align} \int_{-\infty}^\infty\left(\frac1{\alpha+ix}+\frac1{\alpha-ix}\right)^2\mathrm{d}x &=-i\int_\gamma\left(\frac1{\alpha+z}+\frac1{\alpha-z}\right)^2\mathrm{d}z\tag{7}\\ &=-i\int_\gamma\left(\frac1{(\alpha+z)^2}+\frac1{(\alpha-z)^2}\right)\mathrm{d}z\\ &-\frac i\alpha\int_\gamma\left(\color{#C00000}{\frac1{\alpha+z}}+\frac1{\alpha-z}\right)\mathrm{d}z\tag{8}\\ &=\frac{2\pi}{\alpha}\tag{9} \end{align} $$ Explanation:
$(7)$: the integral along the arc where $|z|=R$ is $\lesssim\frac{4\pi}R$, which vanishes as $R\to\infty$.
$(8)$: $\left(\frac1{\alpha+z}+\frac1{\alpha-z}\right)^2 =\frac1{(\alpha+z)^2}+\frac1{(\alpha-z)^2}+\frac2{(\alpha+z)(\alpha-z)} =\frac1{(\alpha+z)^2}+\frac1{(\alpha-z)^2}+\frac1\alpha\left(\frac1{\alpha+z}+\frac1{\alpha-z}\right)$
$(9)$: only the $\color{#C00000}{\frac1{\alpha+z}}$ term has a pole with non-zero residue inside $\gamma$