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

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)?

• Is the integrand really typed correctly? Should this be, e.g., $\frac{1}{a - ix} + \frac{1}{a + ix}$? – Travis Willse Feb 5 '16 at 13:49
• @Travis Good call, thanks for pointing that out. – Aldon Feb 5 '16 at 13:53
• Have you tried combining the fractions? – user170231 Feb 5 '16 at 13:54
• @user170231 Yes, you'd get $\left(\frac{2 \alpha}{\alpha^2 + x^2}\right)^2$ which is another integral that's hard to integrate, which I'm thinking one that can be solved, again, by a contour integral. – Aldon Feb 5 '16 at 13:59

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*}

• I have done the part where you'll combine the fractions to form $\frac{4 \alpha^2}{(\alpha^2 + x^2)^2}$ but I have not thought of factoring out $\alpha^2$ from the denominator. I will try that out thanks! – Aldon Feb 5 '16 at 14:02
• I'm beginning to see that this can be done using $x = \alpha \tan \theta$ but I keep on getting extremely weird answers. – Aldon Feb 5 '16 at 15:10
• Nothing weird Aldon! I completed the answer. I hope it could be clearer now! :-) – Joe Feb 6 '16 at 1:06
• I realized a few hours ago that I had wrong bounds for my integral when I converted it through $x = \alpha \tan \theta$. Thanks a lot though! – Aldon Feb 6 '16 at 1:14
• You're welcome! :-) – Joe Feb 6 '16 at 1:22

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 $\gamma$

• Why is $\left(\frac{1}{\alpha +z} + \frac{1}{\alpha - z}\right)^2 = \frac{1}{(\alpha + z)^2} + \frac{1}{(\alpha - z)^2}$? – Aldon Feb 6 '16 at 1:07
• you could also use the Fourier transform (harmonic analysis) approach. if $\alpha < 0$ : $$\frac{1}{\alpha+ix} + \frac{1}{\alpha-ix} = 2 \pi \int_{-\infty}^\infty e^{-2 i \pi x t} e^{2 \pi \alpha |t|} dt$$ so by the Parseval identity : $$\int_{-\infty}^\infty \left| \frac{1}{\alpha+ix} + \frac{1}{\alpha-ix} \right |^2 dx = \int_{-\infty}^\infty |2 \pi e^{2 \pi \alpha |t|}|^2 dt = 4 \pi^2 \int_{-\infty}^\infty e^{4 \pi \alpha |t|} dt = 2\frac{4 \pi^2 }{4 \pi |\alpha|} = \frac{2 \pi }{|\alpha|}$$ – reuns Feb 6 '16 at 1:45
• and if $\alpha > 0$ : $\frac{1}{\alpha+ix} + \frac{1}{\alpha-ix} = - \left(\frac{1}{-\alpha-ix} + \frac{1}{-\alpha+ix} \right)$ so the result is still $\frac{2 \pi }{|\alpha|}$ – reuns Feb 6 '16 at 1:45
• @user1952009: indeed, there are many ways to compute the integral. – robjohn Feb 6 '16 at 7:03
• @Aldon: I have expanded the explanation of the Complex Analysis Approach. Note the explanation of $(8)$. – robjohn Feb 6 '16 at 7:39