# Improper integrals are “not totally Improper”

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

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

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

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

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

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

$$=\frac{\pi}{2a^3}$$

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 $\int_{\mathcal{T}_R}\frac{dx}{(x^2+a^2)^2}\rightarrow 0$

So,

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

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

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

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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

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.