Thread: Evaluating a real integral

1. Evaluating a real integral

The aim is to evaluate $\displaystyle \int \frac {sin^{2}x}{x^2}dx$. The strategy is to do a contour integral of the function $\displaystyle f(z)=\frac{e^{2iz}-1}{z^{2}+a^{2}}$ for real $\displaystyle a>0$ around the contour $\displaystyle \gamma$ consisting of two parts: $\displaystyle \gamma_{1}$ the semi-circle {$\displaystyle z \in \mathbb{C} | |z|=R, Im(z)>0$}, and $\displaystyle \gamma_{2}$ the straight line segment from $\displaystyle -R$ to $\displaystyle R$.

(i) Explain why $\displaystyle \int_{\gamma_{1}}f(z)dz \to 0$ as $\displaystyle R \to \infty$.

$\displaystyle \int_{\gamma_{1}}f(z)dz=\int_{0}^{\pi}f(\gamma_{1} (t))\gamma_{1} '(t)dt$$\displaystyle =\int_{0}^{\pi} \frac{e^{2iRe^{it}}-1}{R^{2}e^{2it}+a^{2}} iRe^{it} dt$

Now this is where I get stuck.

(ii) Use Cauchy's integral formula to show that $\displaystyle \int_{\gamma}f(z)dz=\frac{\pi}{a}(e^{-2a}-1)$.

Now I've tried to use partial fractions so that $\displaystyle \frac{e^{2iz}-1}{z^{2}+a^{2}}=\frac{e^{2iz}-1}{(z+ai)(z-ai)}=\frac{A}{(z+ai)}+\frac{B}{(z-ai)}=\frac{A(z-ai)+B(z+ai)}{(z+ai)(z-ai)}$

Equating the numerators gives $\displaystyle e^{2iz}-1=A(z-ai)+B(z+ai)$ but I'm not sure how to solve this equation for A and B.

(iii) Show that $\displaystyle e^{2ix}-1=-2sin^{2}x+isin2x$ and hence that $\displaystyle \int_{\gamma_{2}}f(z)dz$ is real.

2. For (i):

$\displaystyle \left|\int_{0}^{\pi} \frac{e^{2iRe^{it}}-1}{R^{2}e^{2it}+a^{2}} iRe^{it} dt\right|\leq \int_0^{\pi} \left|\frac{R(e^{2iRe^{it}}-1)}{R^2e^{2it}-1}\right|dt$

$\displaystyle \leq \int_0^{\pi}\left| \frac{e^{2iR(\cos(t)+i\sin(t)}-1}{R(e^{it}-\frac{1}{R^2})}\right|dt$

and as $\displaystyle R\to\infty$, the integral tends to the form of:

$\displaystyle \frac{1}{R}\int_0^{\pi}e^{-2R\sin(t)}dt$

which tends to zero for unbounded R.

That however may need a little cleaning up but it's a good start I think.

For (ii), I don't see why you doing all that. Just use Cauchy's Integral Formula directly:

$\displaystyle \oint \frac{e^{2iz}-1}{z+ia}\frac{1}{z-ia}dz=2\pi i\left( \frac{e^{2iz}-1}{z+ia}\right)_{z=ia}=\frac{\pi}{a}(e^{-2a}-1)$

For (iii), $\displaystyle \sin(x)$ is an odd function so the imaginary component from (-k,k) is zero right?