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Methods of evaluating $$\int_0^{\infty}\frac{{\rm d}x}{x^2+1}$$

Firstly i know that directly: $$\int_0^{\infty}\frac{{\rm d}x}{x^2+1}=\arctan x\Bigg|_{0}^{\infty}=\frac{\pi}2$$ Also we can use the contour integral: $$\int_{\text{Line}}\frac{{\rm d}z}{z^2+1}+\underbrace{\int_{\text{Arc}}\frac{{\rm d}z}{z^2+1}}_{\text{zero}}=2\pi {\rm i}\;\substack{\large \text{Res}\\{z=i}}\;\frac{1}{z^2+1}=2\pi{\rm i}(1/(2i))=\pi$$ Since the line extends in both directions but we need only the positive side $$\int_0^{\infty}\frac{{\rm d}x}{x^2+1}=\frac{\pi}2$$ Now I have two questions:

  • What methods can be used to evaluate the residue, can someone be detailed enough to explain in brief all useful? Note that I have too listed them below. Actually explain for the function $$\frac1{(z^2+1)^2}$$ at $z=i$
  • What other methods for evaluating the integral?

Residue Methods I would use:
Laurent series, Some theorem which says residue for $\frac{\phi(z)}{(z-z_0)^m}$ at $z_0$ is $\frac{\phi^{(m-1)}(z_0)}{(m-1)!}$, for pole of order 1 in a zero of the denominator of rational function we can diffrentiate the denominator once.

dustin
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RE60K
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  • So you want to see the evaluation of $$\int_0^{\infty}\frac{1}{(z^2 + 1)^2}dz$$ via residue theorem? – dustin Dec 28 '14 at 06:04
  • @dustin I can do it, if you wish may i include it in the question? anyways i meant "different methods to evaluate". – RE60K Dec 28 '14 at 06:05
  • @ADG you don't need to include it. I was just confused at what you were looking for. – dustin Dec 28 '14 at 06:05

2 Answers2

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Let be $$ I_n=\int_0^{\infty}\frac{\mathrm dx}{(1+x^2)^n} $$ We have $$\begin{align} I_n &= \underbrace{\int_0^{\infty}\frac{1+x^2}{(1+x^2)^n}\mathrm dx}_{I_n-1}-\int_0^{\infty}x\cdot \frac{x}{(1+x^2)^n}\mathrm dx\\ &=I_{n-1}-\underbrace{\left.\frac{x}{2(1-n)(x^2+1)^{n-1}}\right|_0^{\infty}}_{\to 0}+\frac{1}{2(1-n)}I_{n-1} \end{align} $$ and finally we obtain the recurrence relation

$$ I_n=\frac{2n-3}{2(n-1)}I_{n-1} \qquad (n>1) $$ with $$ I_1=\int_0^{\infty}\frac{\mathrm dx}{1+x^2}=\frac{\pi}{2} $$

For $n=2$, you have $I_2=\frac{\pi}{4}$.

alexjo
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You can use beta function to evaluate the integral. Using the substitution $t=\frac{1}{1+x^2}$ casts the integral to

$$ I = \int_{0}^{\infty} \frac{1}{(1+x^2)^2}dx = \frac{1}{2} \int_{0}^{1} t^{1/2}\,(1-t)^{-1/2} dt = \beta( 3/2,1/2 )=\frac{\pi}{4}. $$

Note:

$$ \beta( 3/2,1/2 ) = \frac{\Gamma(3/2)\Gamma(1/2)}{\Gamma(2)} = \frac{1}{2}\Gamma(1/2)^2 = \frac{\pi}{2}. $$

Mhenni Benghorbal
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