Residue Theorem and Real integrals

**mia25** · May 17th 2011, 03:04 AM

Hello,

I need help in solving the following real integral using the residue theorem.
$\int_0^\infty \frac{sin^2 x}{x^2}dx$ .
What starting complex function should I use? and what contour?

Thank you

**FernandoRevilla** · May 17th 2011, 07:06 AM

One way:

(a) Consider $\int_C\dfrac{e^{iz}dz}{z},\quad C=\gamma_1\cup\gamma_2\cup \gamma_3\cup\gamma_4$

$\begin{Bmatrix}\gamma_1(t)=t,\;t\in[ \epsilon,R]\\\gamma_2(t)=Re^{it},\;t\in [0,\pi]\\\gamma_3(t)=t,\;t\in [-R,- \epsilon]\\ \gamma_4(t)= \epsilon e^{(\pi-t)i},\;t\in [0,\pi]\end{matrix}$

Using the residue theorem you'll obtain $\int_0^{+\infty}\dfrac{\sin x \;dx}{x}=\dfrac{\pi}{2}$

(b) Using the integration by parts method, you'll obtain

$\int_0^{+\infty}\dfrac{\sin^2 x \;dx}{x^2}=\int_0^{+\infty}\dfrac{\sin 2x \;dx}{x}$

(c) Using the substitution $t=2x$ you'll obtain

$\int_0^{+\infty}\dfrac{\sin 2x \;dx}{x}=\int_0^{+\infty}\dfrac{\sin t \;dt}{t}=\dfrac{\pi}{2}$

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