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I want to know how to solve this using contour integration:

$$\int_0^{\infty} \frac{\sin(x)}{\sqrt{x}}dx.$$

So I let the integral become:

$$\oint_c \frac{\sin(z)}{\sqrt{z}}dz$$ where c is a "half doughnut" shape avoiding the singularity at z = 0 and extending into the upper half of the complex plane towards infinity.

$$\oint_c = \int_{up} + \int_{-R}^{- \epsilon} + \int_{low} + \int_ {\epsilon}^R = 0$$ (Because no singularities are actually contained within the contour.)

By a bound argument, the $\int_{up}$ contributes nothing to the integral. Therefore:

$$\lim{R \to \infty}, {\epsilon \to 0}$$

$$- \int_{low} = \int_{- \infty}^{\infty} $$

Where $\int_{low}$ is the integral over the bump going over the point z = 0. So can I use the Cauchy Integral theorem to say

$$\int_{low} = \pi i\, \text{Res} \left( \frac{\sin x}{ \sqrt{x}}, 0 \right)$$

Because there is no residue for this function, which would imply the integral is zero, which I know it is actually $\sqrt \frac{ \pi}{2}$.

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3 Answers 3

You bounding argument for $\int_{up}$ will not work.

$\sin z$ can be rewritten as $\frac{e^{iz} - e^{-iz}}{2\pi i}$. Over the upper half plane, $|e^{iz}| \to 0$ and $|e^{-iz}| \to \infty$ when $|z| \to \infty$. You will not have any control of your $\int_{up}$ as you send it to $\infty$. To compute this integral, you should:

1) Change variable to $x = \sqrt{z}$ to get rid of the square root.
2) Express $\sin x^2$ in terms of imaginary part of $e^{ix^2}$ to make the integrand better behaved over upper half plane.

$$\int_{0}^{\infty} \frac{\sin{z}}{\sqrt{z}} dz = 2 \int_{0}^{\infty} \sin x^2 d x = 2 \Im\left[\int_{0}^{\infty}e^{ix^2} dx\right]$$

3) Pick a right contour to compute the integral. To pick the right contour, one thing you need to do is understand the behavior of your integrand in various limit.

Let's $x$ = $R e^{i\theta}$ for large $R$ and $\theta \in [0,\frac{\pi}{2}]$, we have:

$$e^{ix^2} = e^{iR^2 \exp(2i\theta)} = e^{-R^2 \sin(2\theta)} e^{iR^2 \cos(\theta)}$$

Notice the $e^{-R^2 sin(2\theta)}$ factor there. It means as long as we are within the $1^{st}$ quadrant, $e^{ix^2}$ will not cause any real problem as you send $|x| \to \infty$.

Is there a contour one can use in $1^{st}$ quadrant? The answer is yes. Let $C$ be the contour which start from $0$ to $R$ on real axis, followed by a circle arc from $R$ to $R e^{i\frac{\pi}{4}}$ and then a straight line from $R e^{i\frac{\pi}{4}}$ back to $0$.

Along this contour, you have:

$$ 0 = \left[\int_0^R + \underbrace{\int_R^{Re^{i\frac{\pi}{4}}}}_{\to 0 \text{ as } R \to \infty} + \int_{Re^{i\frac{\pi}{4}}}^0 \right] e^{ix^2} dx$$

Substitute variable $x = y e^{i\frac{\pi}{4}}$ in last integral, we get:

$$\begin{align} & \int_0^{\infty} e^{ix^2} dx = -e^{i\frac{\pi}{4}} \int_{\infty}^0 e^{-y^2} dy\\ \implies & \int_{0}^{\infty} \frac{\sin{z}}{\sqrt{z}} dz = 2 \Im\left[ e^{i\frac{\pi}{4}} \int_0^{\infty} e^{-y^2} dy \right] = \sqrt{2}\frac{\sqrt{\pi}}{2} = \sqrt{\frac{\pi}{2}} \end{align}$$

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There is a branch cut at zero in what you have.

Here's another, more straightforward approach:

$$ \int_0^\infty\frac{e^{ix}}{\sqrt{x}}\,dx =\sqrt{i\,}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,dx =\frac{1+i}{\sqrt{2}}\Gamma\left(\frac12\right) =(1+i)\sqrt{\frac\pi2}, $$ (where in the second equality we have appealed to the Gamma function).

Now, since since $e^{i x}=\cos x+i\sin x$, we have $\sin x=\text{Im}(e^{ix})$. Thus, taking the imaginary part, $$\int_0^\infty \frac{\sin x}{\sqrt x}\,dz=\sqrt{\frac{\pi}{2}}.$$

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1  
A complex substitution transforms a definite integral into a contour integral on the complex plane. So what you wrote, while true, needs justification. –  Random Variable Feb 17 '13 at 5:52

\begin{align} \int_{0}^{\infty}{\sin\left(x\right) \over \sqrt{x\,}}\,{\rm d}x &= 2\int_{0}^{\infty}\sin\left(x^{2}\right)\,{\rm d}x = 2\Im\left\lbrack% \int_{0}^{\infty}{\rm e}^{{\rm i}x^{2}}\,{\rm d}x \int_{0}^{\infty}{\rm e}^{{\rm i}y^{2}}\,{\rm d}y \right\rbrack^{1/2} \\[3mm]&= 2\,\sqrt{\pi \over 2\,}\ \Im\left\lbrack% \int_{0}^{\infty}{\rm e}^{{\rm i}\,\rho^{2}}\rho\,{\rm d}\rho\, \right\rbrack^{1/2} = \sqrt{\pi\,}\ \Im\left\lbrack% \int_{0}^{\infty}{\rm e}^{{\rm i}\,\rho}\,{\rm d}\rho\, \right\rbrack^{1/2} \\[3mm]&= \sqrt{\pi\,}\ \Im\left\lbrace{\rm i}% \int_{-\infty}^{\infty}\left\lbrack \int_{-\infty}^{\infty}{{\rm d}k \over 2\pi}\,{{\rm e}^{-{\rm i}k\rho} \over k + {\rm i}0^{+}}\right\rbrack {\rm e}^{{\rm i}\,\rho}\,{\rm d}\rho\, \right\rbrace^{1/2} \\[3mm]&= \sqrt{\pi\,}\ \Im\left\lbrace{\rm i}% \int_{-\infty}^{\infty}{{\rm d}k \over k + {\rm i}0^{+}}\left\lbrack \int_{-\infty}^{\infty}{\rm e}^{-{\rm i}\,\rho\left(k - 1\right)}\,{{\rm d}\rho \over 2\pi} \right\rbrack \right\rbrace^{1/2} \\[3mm]&= \sqrt{\pi\,}\ \Im\left\lbrack{\rm i}% \int_{-\infty}^{\infty}{{\rm d}k \over k + {\rm i}0^{+}}\,\delta\left(k - 1\right) \right\rbrack^{1/2} = \sqrt{\pi\,}\ \Im\left({{\rm i} \over 1 + {\rm i}0^{+}}\right)^{1/2} \\[3mm]&= \sqrt{\pi\,}\ \Im\left\lbrack{{\rm i} + \pi\,\delta\left(1\right)}\right\rbrack^{1/2} = \sqrt{\pi\,}\ \Im\left(\,{\rm i}\,\right)^{1/2} = \sqrt{\pi\,}\ \Im\left({\rm e}^{{\rm i}\,\pi\,/\,2}\right)^{1/2} = \sqrt{\pi\,}\ \Im{\rm e}^{{\rm i}\,\pi\,/\,4} \\[3mm]&= \sqrt{\pi\,}\ \sin\left(\pi \over 4\right) = \sqrt{\pi \over 2} \end{align}

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The use of the delta "function" will need some justification since the integral used to get it does not converge. –  robjohn Aug 22 '13 at 23:59

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