# How can we show that $\int_{-\infty}^{+\infty}{b\over 1+2ax+(cx)^2}\cdot{\mathrm dx\over 1-2ax+(cx)^2}={\pi\over 2}?$

Consider this integral

$$\int_{-\infty}^{+\infty}{b\over 1+2ax+(cx)^2}\cdot{\mathrm dx\over 1-2ax+(cx)^2}={\pi\over 2}\tag1$$ Where $a^2+b^2=c^2$; Pythagoras theorem.

Making an attempt:

Complete the square of $(cx)^2+2ax+1=c^2\left(x+{a\over c^2}\right)^2+{c^2-a^2\over c^2}$, I can't see how this would be of any useful.

$1+2x^2(c^2-2a^2)+c^4x^4=(1+2ax+c^2x^2)(1-2ax+c^2x^2)$

I am even sure how to go about making an attempt to find the equivalent of $(1)$.

How may we go about and prove $(1)?$

• Have you tried expressing the integrand in partial fractions? – David Quinn Apr 24 '17 at 21:09
• That would be long, right? I did thought about it. – gymbvghjkgkjkhgfkl Apr 24 '17 at 21:11
• a very basic application of the residue theorem will do the job – tired Apr 24 '17 at 23:15

Let $I$ denote the integral and write $(a, b) = (c\alpha, c\beta)$ so that $\alpha^2 + \beta^2 = 1$. Then by the substitution $x = 1/ct$, we have

\begin{align*} I = \int_{-\infty}^{\infty} \frac{b}{(1 + c^2x^2)^2 - 4a^2x^2} \, dx &= \int_{-\infty}^{\infty} \frac{\beta}{(t + t^{-1})^2 - 4\alpha^2} \, dt \\ &= \int_{-\infty}^{\infty} \frac{\beta}{(t - t^{-1})^2 + 4\beta^2} \, dt. \end{align*}

Now it follows from the Glasser's master theorem1) that

$$I = \int_{-\infty}^{\infty} \frac{\beta}{u^2 + 4\beta^2} \, du = \left[ \frac{1}{2}\arctan\left(\frac{u}{2\beta}\right) \right]_{-\infty}^{\infty} = \frac{\pi}{2}.$$

$\text{1)}$ Notice that the following good old trick also works: substituting $t \mapsto 1/t$, we have

$$\int_{0}^{\infty} \frac{\beta}{(t - t^{-1})^2 + 4\beta^2} \, dt = \int_{0}^{\infty} \frac{t^{-2} \beta}{(t - t^{-1})^2 + 4\beta^2} \, dt.$$

Thus it follows that

$$I = \int_{0}^{\infty} \frac{(1 + t^{-2}) \beta}{(t - t^{-1})^2 + 4\beta^2} \, dt = \int_{-\infty}^{\infty} \frac{\beta}{u^2 + 4\beta^2} \, du$$

where $u = t - 1/t$. The rest is the same.

I will assume $b > 0$ (in which case $a^2$ can never equal $c^2$).

Employing a partial fraction decomposition we have $$\frac{1}{1 + 2ax + c^2 x^2} \cdot \frac{1}{1 - 2ax + c^2 x^2} = \frac{1}{4a} \cdot \frac{2a + c^2 x}{1 + 2ax + c^2 x^2} + \frac{1}{4a} \cdot \frac{2a - c^2 x}{1 - 2ax + c^2 x^2}.$$ Now noting that $$\int \frac{2a + c^2 x}{1 + 2ax + c^2 x^2} \, dx = \frac{a}{\sqrt{c^2 - a^2}} \tan^{-1} \left (\frac{c^2 x + a}{\sqrt{c^2 - a^2}} \right ) + \frac{1}{2} \ln |1 + 2ax + c^2 x^2| + \cal{C}_1,$$ and $$\int \frac{2a - c^2 x}{1 - 2ax + c^2 x^2} \, dx = \frac{a}{\sqrt{c^2 - a^2}} \tan^{-1} \left (\frac{c^2 x - a}{\sqrt{c^2 - a^2}} \right ) - \frac{1}{2} \ln |1 - 2ax + c^2 x^2| + \cal{C}_2.$$ (note that each of these integrals can be solved by elementary means). Thus \begin{align*} \int_{-\infty}^{\infty}\frac{b}{1+2ax+c^2x^2}\cdot \frac{dx}{1-2ax+c^2 x^2} &= \left [\frac{b}{4 \sqrt{c^2 - a^2}} \left \{\tan^{-1} \left (\frac{c^2 x + a}{\sqrt{c^2 - a^2}} \right ) \right. \right.\\ & \left. \left. + \tan^{-1} \left (\frac{c^2 x - a}{\sqrt{c^2 - a^2}} \right ) \right \}+ \frac{b}{8a} \ln \left |\frac{1 + 2ax + c^2 x^2}{1 - 2ax + c^2 x^2} \right | \right ]^\infty_{-\infty}\\ &= \frac{b}{4 \sqrt{c^2 - a^2}} \left [\frac{\pi}{2} + \frac{\pi}{2} - \left (-\frac{\pi}{2} - \frac{\pi}{2} \right ) \right ]\\ &= \frac{\pi b}{2 \sqrt{c^2 - a^2}}. \end{align*} But given $a^2 + b^2 = c^2$, then $\sqrt{c^2 -a^2} = \sqrt{b^2} = |b| = b$, assuming $b > 0$ giving $$\int_{-\infty}^\infty \frac{b}{1+2ax+c^2x^2} \cdot \frac{dx}{1-2ax+c^2 x^2} = \frac{\pi}{2},$$ as required to show