1. ## Calc of Residues

Hi! I've learned how to take contour integrals for functions that require branch cuts to find improper integrals. However, is there a way to find, say, $\int_0^{\infty}\frac{\sin x}{\sqrt{x}}dx$ using a similar method? I haven't seen any examples involving trig functions with multifunctions.

Thanks for any help!

2. Yes, we can do this with CA. I am skiddish about doing this. Hope I don't make a mistake somewhere.

$\int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx$

Make the sub $u=\sqrt{x},\;\ u^{2}=x$

Then, we can write it as $2\int_{0}^{\infty}sin(u^{2})du$

Try using a 'wedge' in the 1st quadrant of the complex plane with angle

${\phi}=\frac{\pi}{4}$

Now, using $f(z)=e^{z^{2}}$

By the residue theorem, $\int e^{-z^{2}}dz=2{\pi}i\sum\text{enclosed residues}$

Because there are no singularities, the enclosed residue = 0.

Since the wedge is broken up into three different sections along the contour, we can break it up into three separate integrals. Call them a,b, and c for instance.

$\int e^{-z^{2}}dz=\int_{a} e^{-z^{2}}dz+\int_{b}e^{-z^{2}}dz+\int_{c}e^{-z^{2}}dz$

Take the radius of the wedge to be fixed for now at r=R and use the polar:

$z=re^{i{\theta}}$

$dz=dr(e^{i{\theta}})+ire^{i{\theta}}$

First, take a peek at curve a, which lies on the x-axis.

Along this curve, $d{\theta}=0, \;\ {\theta}=0, \;\ z=r, \;\ dz=dr$

Curve b, is circular in its path at radius R, so while ${\theta}$ varies, r is fixed. So, along b, $r=R, \;\ dr=0, \;\ dz=iRe^{i{\theta}}d{\theta}$.

Next, along c, ${\theta}$ is fixed. But this time ${\theta}=\frac{\pi}{4}, \;\ d{\theta}=0, \;\ dz=e^{i\frac{\pi}{4}}dr$

So, the integral from above becomes:

$\int e^{-z^{2}}dz=\int_{0}^{R}e^{-z^{2}}dr+\int_{0}^{\frac{\pi}{4}}e^{-R^{2}e^{i2{\theta}}}iRe^{i{\theta}}d{\theta}+\int_ {0}^{R}e^{-r^{2}e^{i2{\theta}}}e^{\frac{i{\pi}}{4}}dr$

Now, let $R\to {\infty}$. On b, as $R\to {\infty}$,

we get $\lim_{R\to {\infty}}\int_{0}^{\frac{\pi}{4}}e^{-R^{2}e^{i2{\theta}}}iRe^{i{\theta}}d{\theta}=0$

because $e^{-R^{2}e^{i2{\theta}}}\to 0$ is a lot faster than $R\to {\infty}$.

Remember that $\int e^{-z^{2}}dz=0$. Then we are left with

$\lim_{R\to {\infty}}\left[\int_{0}^{R}e^{-r^{2}}dr-\int_{0}^{R}e^{-r^{2}e^{2i{\theta}}}e^{\frac{{\pi}i}{4}}dr\right]=0$

The first one is a familiar one which equals $\int_{0}^{\infty}e^{-r^{2}}dr=\frac{\sqrt{\pi}}{2}$

By Ol' Euler: $e^{\frac{{\pi}i}{4}}=cos(\frac{\pi}{4})+isin(\frac {\pi}{4})=\frac{1+i}{\sqrt{2}}$

$\frac{\sqrt{\pi}}{2}=\frac{1}{\sqrt{2}}\int_{0}^{\ infty}cos(r^{2})dr+\frac{1}{\sqrt{2}}\int_{0}^{\in fty}sin(r^{2})dr+\frac{i}{\sqrt{2}}\int_{0}^{\inft y}[cos(r^{2})dr-sin(r^{2})]dr$

Equate real and imaginary and note that $Im(\frac{\sqrt{\pi}}{2})=0$

$\frac{i}{\sqrt{2}}\int_{0}^{\infty}[cos(r^{2})-sin(r^{2})]dr=0$

and $\int_{0}^{\infty}cos(r^{2})dr=\int_{0}^{\infty}sin (r^{2})dr$

$\frac{\sqrt{\pi}}{2}=\frac{1}{\sqrt{2}}\int_{0}^{\ infty}cos(r^{2})dr+\frac{1}{\sqrt{2}}\int_{0}^{\in fty}sin(r^{2})dr=\frac{2}{\sqrt{2}}\int_{0}^{\inft y}sin(r^{2})dr$

$\int_{0}^{\infty}sin(r^{2})dr=\frac{1}{2}\sqrt{\fr ac{\pi}{2}}$

Therefore, from the top, $\int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx=\sqrt{\ frac{\pi}{2}}$

3. I'm probably going to regret asking this, but why does $e^{-r^{2}e^{2i \theta}}\Big(\frac{1+i}{\sqrt{2}}\Big) = \frac{1}{\sqrt{2}} \Big(\cos r^{2} + \sin r^{2}\Big) + \frac{i}{\sqrt{2}} \Big(\cos r^{2} - \sin r^{2} \Big)$ ?

4. Using these formulas ( obtained by Euler )

$\int_0^{\infty} x^{n-1} e^{-px} \cos(qx) ~dx = \frac{ \Gamma(n) }{ r^n} \cos(n \alpha)$ and

$\int_0^{\infty} x^{n-1} e^{-px} \sin(qx) ~dx = \frac{ \Gamma(n) }{ r^n} \sin(n \alpha)$

$r = \sqrt{ p^2 + q^2 } ~~~ \alpha = \tan^{-1}( \frac{ q}{p} )$