Real integral with complex analysis

**Dinkydoe** · April 23rd 2010, 08:04 AM

I have some trouble evaluating the following integral using complex analysis:

$\int_{0}^{\infty} \frac{\cos(\alpha x)}{x^2}dx$

I believe the trick is to complexify the integrand, let $g(z)=e^{\alpha z}/z^2$ and then make a contour:

$\gamma_1(t) = t, t\in [-R,-1/R]$
$\gamma_2(t)=e^{it}/R,t\in[\pi,2\pi]$
$\gamma_3(t) = t, t\in [1/R,R]$
$\gamma_4(t)= e^{it}, t\in [0,\pi]$

And $\gamma_R = \bigcup \gamma_i$ . Then we can evaluate $\lim_{R\to\infty}\int_{\gamma_R}g(z)dz = 2\alpha\pi i$ with the Cauchy-integral formula.

By Jordan's Lemma $\int_{\gamma_4}g(z)dz=0$ . We like to evaluate $Re(\lim_{R\to\infty}\int_{\gamma_1\cup\gamma_3}g(z )dz) = \int_{-\infty}^{\infty} \frac{\cos(\alpha x)}{x^2}dx$

Can someone tell me how far off I am, and maybe the correct approach for this. My problem is, I have no idea how to evaluate $\int_{\gamma_2}g(z)dz$ , the partial residu around the second order singularity z=0.

So is this the right approach, if not, what is? And if yes, how to evaluate the afore mentioned partial residue.

**chisigma** · April 23rd 2010, 01:32 PM

Let suppose to 'attack' the slightly different integral...

$\int_{0}^{\infty} \frac{1 - \cos (a x)}{x^{2}}\cdot dx$ (1)

At this scope we can compute the complex integral...

$\int_{\Gamma} \frac{1 - e^{i a z}}{z^{2}}\cdot dz$ (2)

... along the red path $\Gamma$ indicated in figure...

... when $r \rightarrow 0$ and $R \rightarrow \infty$ . Since no singularities lie inside the path is...

$\int_{\Gamma} \frac{1 - e^{i a z}}{z^{2}}\cdot dz = 0$ (3)

... and for the Jordan's lemma the integral along the 'big half circle' tends to $0$ if $R \rightarrow \infty$ . Tacking into account that the (3) becomes...

$\int_{-R}^{-r} \frac{1 - e^{i a x}}{x^{2}}\cdot dx + i\cdot \int_{\pi}^{0} \frac{1 - e^{i a r e^{i \theta}}}{r}\cdot e^{-i \theta}\cdot d\theta + \int_{r}^{R} \frac{1 - e^{i a x}}{x^{2}}\cdot dx = 0$ (4)

Now if $r \rightarrow 0$ and $R \rightarrow \infty$ , tacking into account that ...

$\lim_{r \rightarrow 0} i\cdot \int_{\pi}^{0} \frac{1 - e^{i a r e^{i \theta}}}{r}\cdot e^{-i \theta}\cdot d\theta = - a\cdot \pi$ (5)

... from (4) we derive that...

$\int_{0}^{\infty} \frac{1-\cos (a x)}{x^{2}}\cdot dx = \frac{a}{2}\cdot \pi$ (6)

From (6) is evident that is...

$\int_{0}^{\infty} \frac{1}{x^{2}}\cdot dx = \int_{0}^{\infty} \frac{\cos (a x)}{x^{2}}\cdot dx + \frac{a}{2}\cdot \pi$ (7)

... and, because the integral in first member diverges, the integral in second member diverges too...

Kind regards

$\chi$ $\sigma$

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