I need some reassurance that what I did here actually shows what need to be shown. Please correct me if I'm wrong.

In Donald Sarason's "Notes on complex function theory", this question appears at section VII.4, (pg 93) in the context of Cauchy's theorem for convex region. To be accurate, the exercise is

Let $0<b<1$. Derive the equality $$\int_{-\infty}^{\infty}\frac{1-b+x^{2}}{\left(1-b+x^{2}\right)^{2}+4bx^{2}}dx=\pi$$ by integrating the function $\left(1+z^{2}\right)^{-1}$ around the rectangle with vertices $\pm a,\pm a+i\sqrt{b}$ $a>0$ and taking limit as $a\rightarrow\infty$

I tried to follow the suggestion and got (by Cauchy's theorem for convex region):
$$\int_{-a}^{a}\left(1+t^{2}\right)^{-1}dt+\int_{0}^{\sqrt{b}}\left(1+\left(a+it\right)^{2}\right)^{-1}\cdot i\cdot dt+\int_{a}^{-a}\left(1+\left(i\sqrt{b}+t\right)^{2}\right)^{-1}dt+\int_{\sqrt{b}}^{0}\left(1+\left(-a+it\right)^{2}\right)^{-1}\cdot i\cdot dt=0$$

Or in somewhat more conventional form:

Assuming I didn't make any mistakes so far, we get that RHS is zero hence:

Taking limit $a\rightarrow\infty$ gives us $\pi$ on the RHS. So We'll have to figure out what the LHS has to do with the function in the question...

Cauchy's theorem gives us "free of charge" that the integral of any imaginary part will be equal to zero, so in order to simplify the real part, I'll multiply and divide by the complex conjugate of $1+\left(i\sqrt{b}-t\right)^{2} $ which is $\overline{1-b+t^{2}-2i\sqrt{b}t}=1-b+t^{2}+2i\sqrt{b}t$ and get:


hence (by preceding statement) we have:
$$\lim_{a\rightarrow\infty}\int_{-a}^{a}\frac{1-b+t^{2}}{\left(1-b+t^{2}\right)^{2}+4bt^{2}}dt=\lim_{a\rightarrow\infty}\int_{-a}^{a}\left(1+t^{2}\right)^{-1}dt$$ And (as stated before and easy to see): $$\lim_{a\rightarrow\infty}\int_{-a}^{a}\left(1+t^{2}\right)^{-1}dt=\pi$$ So indeed $$\int_{- \infty}^{\infty}\frac{1-b+t^{2}}{\left(1-b+t^{2}\right)^{2}+4bt^{2}}dt = \pi$$


1 Answer 1


If you make the substitution $t =-u$, then $$\begin{align} \int_{\sqrt{b}}^{0} \left(1+(-a+it)^{2} \right)^{-1} \, i \, dt &= - i \int_{-\sqrt{b}}^{0} \left(1+(-a-iu)^{2} \right)^{-1} \, du \\ &= - i\int_{-\sqrt{b}}^{0} \left(1+(a+iu)^{2} \right)^{-1} \, du .\end{align} $$

But the integrand is not an even function in $u$, so$$- i\int_{-\sqrt{b}}^{0} \left(1+(a+iu)^{2} \right)^{-1} \, du \ne - i\int_{0}^{\sqrt{b}} \left(1+(a+iu)^{2} \right)^{-1} \, du.$$

Therefore, the integrals on the right and left sides of the contour don't cancel each other, but they do vanish when you let $ a \to \infty$.

For example, when $a >1$, $$ \left|\int_{0}^{\sqrt{b}} \frac{1}{1+(a+it)^{2}} \, i \, dt \right| \le \int_{0}^{\sqrt{b}} \left|\frac{1}{1+(a+it)^{2}} \right| \, dt \le \sqrt{b} \max_{0 \le t \le \sqrt{b}}\left|\frac{1}{1+(a+it)^{2}} \right| $$ $$\le \sqrt{b} \max_{0 \le t \le \sqrt{b}}\frac{1}{\left|1- \left|a+it \right|^{2} \right|} = \frac{\sqrt{b}}{a^{2}-1} \, , $$ which vanishes as $ a \to \infty$.

Also, if $b>1$, then there is a simple pole inside the contour. And if $b=1$, there is a simple pole on the contour.

So if $b>1$, $$\int_{-\infty}^{\infty}\frac{dt }{1+ (i \sqrt{b}-t)^{2}}=\int_{-\infty}^{\infty}\frac{dt}{1+t^{2}} - 2 \pi i \, \text{Res} \, \left[\frac{1}{1+z^{2}}, i \right] =0 \, , $$ which implies $$\int_{-\infty}^{\infty}\frac{1-b+t^{2}}{\left(1-b+t^{2}\right)^{2}+4bt^{2}} \, dt= 0. $$

And if $b = 1$, $$\text{PV} \int_{-\infty}^{\infty}\frac{dt }{1+ (i -t)^{2}}=\int_{-\infty}^{\infty}\frac{dt}{1+t^{2}} - \pi i \, \text{Res} \, \left[\frac{1}{1+z^{2}}, i \right] = \frac{\pi}{2} \, , $$ which implies $$ \int_{-\infty}^{\infty}\frac{1}{t^{2}+4} \, dt= \frac{\pi}{2}.$$

  • $\begingroup$ I used a linear change of variable $t$ by $-t$, are you sure this is incorrect? $\endgroup$
    – Uria Mor
    Commented May 4, 2016 at 6:51
  • $\begingroup$ You didn't change the limits of integration accordingly. $\endgroup$ Commented May 4, 2016 at 8:16
  • 1
    $\begingroup$ $$ \begin{align} \int_{\sqrt{b}}^{0} \left(1+(-a+it)^{2} \right)^{-1} \, i \, dt &= - i \int_{-\sqrt{b}}^{0} \left(1+(-a-iu)^{2} \right)^{-1} \, du \\ &= - i\int_{-\sqrt{b}}^{0} \left(1+(a+iu)^{2} \right)^{-1} \, du \end{align} $$ But the integrand is not an even function of $u$. So $$- i\int_{-\sqrt{b}}^{0} \left(1+(a+iu)^{2} \right)^{-1} \, du \ne - i\int_{0}^{\sqrt{b}} \left(1+(a+iu)^{2} \right)^{-1} \, du$$ $\endgroup$ Commented May 4, 2016 at 15:14
  • 1
    $\begingroup$ Furthermore, if $0 < b< 1$, the equation $$\int_{-a}^{a}\left(1+\left(i\sqrt{b}-t\right)^{2}\right)^{-1} \, dt=\int_{-a}^{a}\left(1+t^{2}\right)^{-1} \, dt $$ doesn't hold for all $a >0$. It only holds if you let $a \to \infty$. $\endgroup$ Commented May 4, 2016 at 15:18
  • $\begingroup$ I will very much appreciate it (and I think others would too...), if you could edit your answer and add all of these great explanations of yours in the comments to the original answer. $\endgroup$
    – Uria Mor
    Commented May 5, 2016 at 7:57

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