Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question is to evaluate $$\int _{-\infty}^{\infty} \frac{dx}{(x^2+a^2)^2}\text {for } a>0$$

Idea is to calculate this using complex analysis/residue theory/contour integration.

Approach is consider contour $D_R$ consisting of a semicircle in upper half plane of radius $R$ with the line $[-R,R]$

(I am not familiar with idea how to draw figures in latex so, it would be better if some one can help me out if they are sure that they understood what i actually mean).

So, then, we have $$\int_{\partial D_R} \frac{dx}{(z^2+a^2)^2}= \int_{-R}^{R}\frac{dx}{(x^2+a^2)^2}+ \int_{\mathcal{T}_R}\frac{dx}{(x^2+a^2)^2}$$

where $\partial D_R$ is boundary of contour $D_R$ and $\mathcal{T}_R$ is contour except the line $[-R,R]$.

Now, as $D_R$ is bounded domain, we can use residue theorem to find what is $$\int_{\partial D_R} \frac{dx}{(z^2+a^2)^2}$$

we have $$\int_{\partial D_R} \frac{dx}{(z^2+a^2)^2}=\int_{\partial D_R} \frac{dx}{(z+ai)^2(z-ai)^2}$$

$$=2\pi i .\text{Residue at } (ai) $$

$$=2\pi i .\lim_{x\rightarrow ai} \frac{d}{dx}\frac{1}{(z+ai)^2}$$

$$= 2\pi i \lim_{x\rightarrow ai} \frac{-2}{(z+ai)^3}$$

$$=2\pi i \frac{-2}{(2ai)^3}$$

$$=2\pi i\frac{-2}{-8a^3i}$$


So, I have $$\frac{\pi}{2a^3}= \int_{-R}^{R}\frac{dx}{(x^2+a^2)^2}+ \int_{\mathcal{T}_R}\frac{dx}{(x^2+a^2)^2}$$

i.e., $$\int_{-R}^{R}\frac{dx}{(x^2+a^2)^2} = \frac{\pi}{2a^3} - \int_{\mathcal{T}_R}\frac{dx}{(x^2+a^2)^2}$$

as $R \rightarrow \infty $ we see that $\int_{\mathcal{T}_R}\frac{dx}{(x^2+a^2)^2}\rightarrow 0$


$$\int _{-\infty}^{\infty} \frac{dx}{(x^2+a^2)^2}=\frac{\pi}{2a^3}$$

Now, I would be thankful if some one can help me what i have done is valid and I am afraid this should be the case always at least when considering $\int_{\mathcal{T}_R}\frac{dx}{f(x)}$ for $f(x)$ a polynomial

What exactly i mean is we do not have to bother about any other extra conditions except residue theorem when considering $$\int _{-\infty}^{\infty} \frac{dx}{f(x)}$$ because in any case i am fixing a bound for $\int_{\mathcal{T}_R}\frac{dx}{f(x)}$ which goes to $0$ as $R\rightarrow 0$

So, what i would like to say is $\int_{\mathcal{T}_R}\frac{dx}{f(x)}$ is actually seen as $\int _{\partial D_R}$ where $R$ is maximum magnitude of zeros of $f(x)$ in upper half plane.

I am a bit afraid if i am missing some thing.

I would like someone to verify if my idea is true.

$$\int_{\mathcal{T}_R}\frac{dx}{f(x)}=2\pi i \sum {\text{Res. at zeros of f(x)}}$$

If this is the case always then I would like to say

$$"\text{In contrast to its name, Improper Integrals behave properly (conditions apply)}"$$

share|cite|improve this question
Someone else will hopefully confirm, but yes, you should be able to apply the Residue Thm quite readily in the case of polynomials. – Bennett Gardiner Oct 31 '13 at 12:08
I hope the same :) Thankyou @BennettGardiner – Praphulla Koushik Oct 31 '13 at 12:25
up vote 2 down vote accepted

It seems that the only problem you need to worry about is the integral over $ \mathcal{T}_R$, otherwise the approach clearly works. If the polynomial has degree $d = \deg f \geq 2$, then you can write $f(x) = a_0 x^d + a_1x^{d-1}+\dots = \Theta(x^d)$, where by this notation I mean that there are constants $r,C_1,C_2$ such that if $|x|>r$ then $C_1|x|^d< |f(x)|<C_2|x|^d$. Thus, the integral $\int_{\mathcal{T}_R} \frac{dx}{f(x)}$ can be estimated by: $$ \left| \int_{\mathcal{T}_R} \frac{dx}{f(x)}\right| \leq \frac{\text{length of $ \mathcal{T}_R$}}{\text{maximum of } |f(x)|} \leq \frac{\pi R}{C_1 R^d} = \frac{\pi }{C_1 }\cdot \frac{1}{R^{d-1}}.$$ This obviously tends to $0$ with $R \to \infty$ so you can be sure this term can be omitted in the limit.

share|cite|improve this answer
I was thinking of similar idea but was not so sure how to prove... Thank You... – Praphulla Koushik Oct 31 '13 at 12:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.