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Context: I am trying to show that $\int_0^\infty x^{-1}\sin x dx = \frac{\pi}{2}$ using complex analysis, by first integrating $\oint_{\Gamma} z^{-1}e^{iz}$, where $\Gamma$ is a closed contour consisting of two portions of real axis $-R$ to $\epsilon$ and $\epsilon$ to $R$ and two semicircles $\Gamma_1(0,R)$ and $\Gamma_2(0,\epsilon)$.

I am almost done. I just need to show that $$\lim_{R\to\infty}\int_{\Gamma_1} \frac{e^{iz}}{z} = 0.$$

I don't need a rigorous proof. I am new to complex analysis and know almost no theorems. I just want to understand things "intuitively", if it's possible. So far, I know only Cauchy formula/theorem and residue theorem.

I found following solution, which I don't understand very well:

Integrate by parts $\int_{\Gamma_1} \frac{e^{iz}}{z} = \frac{e^{iz}}{iz}|_{-R}^{R} + \int_{\Gamma_1}\frac{e^{iz}}{z^2}dz$. The first turns out to be $\frac{2i}{R}\sin R \to 0$. Then absolute value of the integral is bounded by $\pi R \max_{\Gamma_1}{|\frac{e^{iz}}{z^2}|}\leq\frac{\pi}{R} \to 0$.

Why does it suffice to evalute at $R$ and $-R$? Why is the integral bounded by this?

Thank you!

mirgee
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    I don't know how to put it well. The formula for integration by parts for path integrals is $$\int_\gamma f(z)g'(z),dz = f(\gamma(b))g(\gamma(b)) - f(\gamma(a))g(\gamma(a)) - \int_\gamma f'(z)g(z),dz,$$ if $[a,b]$ is the parameter interval of $\gamma$. If you plug in the definition of the path integral $$\int_\gamma h(z),dz = \int_a^b h(\gamma(t))\cdot\gamma'(t),dt,$$ it reduces to the ordinary integration by parts for real intervals familiar from calculus. So the gist is: You evaluate at the two endpoints of the path. The path $\Gamma_1$ goes from $R$ to $-R$. – Daniel Fischer Jul 25 '14 at 10:57

2 Answers2

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One uses here the inequalilty $$\left|\int_\Gamma f(z)\,dz\right|\leq|\Gamma|\max\limits_{z\in\Gamma}|f(z)|$$ ($|\Gamma|$ being the length of path $\Gamma$), which is almost obvious from the definition of integral of complex variable.

In your case $|\Gamma_1|=\pi R$ (half of a circle of radius $R$) and $|\frac{e^{iz}}{z^2}|=\frac{1}{|z|^2}=\frac{1}{R^2}$ for all $z\in\Gamma_1$, so $\max\limits_{z\in\Gamma_1}|\frac{e^{iz}}{z^2}|=\frac{1}{R^2}$.

CuriousGuest
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For $z=Re^{i\theta}$, $0\leq \theta\leq\pi $, $$\left| e^{i z}/z\right|\leq e^{-R}/R.$$ Use this to bound the original integral. No need for anything fancier.

zibadawa timmy
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