Complex Integration - Poles on the Imaginary axis

[knowlewj01]

Homework Statement

evaluate the integral:
$I_1 =\int_0^\infty \frac{dx}{x^2 + 1}$

by integrating around a semicircle in the upper half of the complex plane.

Homework Equations

The Attempt at a Solution

first i exchange the real vaiable x with a complex variable z & factorize the denominator. Also, the contour of integration is a semicircle with radius= infinity

$I_2 = \int_{-\infty}^{\infty}\frac{dz}{(z+i)(z-i)}$

the contour contains only the pole in the upper half, so from residue theorem we know:
$I_2 = 2\pi i R(i)$

where R(i) is the residue at the point z=i

R(i) = 1/(i+i) = 1/2i

Hence $I_2 = \pi$

Now, i know the answer to the original integral is supposed to be pi/2.
Can i say that: because the original limits range from 0 to infinity, and i have integrated twice this amount, my answer should be divided by 2? or is this reasoning flawed?

 

[Whovian]

I THINK you do divide by two, as noted, but how I'd approach this would be by using trig substitution, substituting x = tan(θ). :P But yep, I think your reasoning's right, or at least close!

 

[jackmell]

Yes.

 

[knowlewj01]

Thanks for the replies. Is this because the function is even in the upper half of the complex plane?

I thought of doing this by integrating a contour in only the positive quadrent, ie:
(0,0) to (R,0)
(R,0) to (0,iR) along contour ω [a radial path of radius R from the real axis to the imaginary axis in the positive quadrent]
(0,iR) to (0,i+δ) [where δ is a small value which we will take to 0 around the pole]
(0,i+δ) to (0,i-δ) along a contour λ [radial path of radius δ in the clockwise direction such that the pole is not included in the enclosed path]
(0,i-δ) to (0,0)

we have:

$F(z) = \frac{1}{Z^2+1}=\frac{1}{(z+i)(z-i)}$

$\oint_C F(z)dz =\int_0^R F(z)dz + \int_\omega F(z)dz + \left[\int_{iR}^{i+\delta}F(z)dz+\int_{i-\delta}^0 F(z)dz\right] + \int_\lambda F(z)dz =0$

as F(z) is analytic everywhere inside the contour we demand that the integral = 0 by Green's theorem.

we know from residue theorem that integration on the contour λ will give a contribution of -πiR(i) as it is counterclockwise (& where R(i) is the residue at i, which is 1/2i)

now let R→∞ and δ→0:

First: consider the integral along the contour ω, if we have $z=Re^{i\theta}$

$\int_\omega F(z)dz = \int_\omega \frac{iRe^{i\theta}d\theta}{R^2e^{i2\theta} + 1}$

$\lim_{R \to \infty}\int_\omega \frac{iRe^{i\theta}d\theta}{R^2e^{i2\theta}+1}≈ \frac{1}{R}\int_\omega \frac{id\theta}{e^{i\theta}}=0$

Is this correct? I thought Jordan's Lemma worked only for semicircles but this seems to give the same result.

we are left with:

$\int_0^\infty F(z)dz + \int_{i\infty}^0 F(z)dz - \frac{\pi}{2} = 0$

so i know that the integral along the imaginary axis must be 0 but I'm not sure how to prove it