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I am trying to use contour Integration to show $$\int_0^\infty \frac{\sin(ax)}{x}=sign(a)\frac{\pi}{2}$$

I was reading a similar example here of $$\int_0^\infty \frac{\sin(x)}{x}=\frac{\pi}{2}$$

I am trying to think how a negative or positive $a$ would change the problem.

1) Is there a way to not do the integral in two cases? One negative and the other positive a?

2) If the a is negative, do I need to make the contour a semicircle in the third and fourth quadrants instead of the first and second quadrants like in the example?

Thank you for your help and time.

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Just notice that $\sin$ is a odd function. If $a<0$ then we have $$\int_0^\infty \frac{\sin(ax)}{x}dx = -\int_0^\infty \frac{\sin(-ax)}{x}dx.$$

So, you just need to calculate the integral when $a>0$. You may also suppose $a=1$ making a change of variable.

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    $\begingroup$ Thank you so much :) $\endgroup$ – MathIsHard Jul 14 '18 at 2:11
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    $\begingroup$ It does not depend because of the following calculation. If $a>0$ then taking $y=ax$ we obtain $$\int_0^\infty \frac{\sin(ax)}{x}dx =\int_0^\infty \frac{\sin(ax)}{ax}adx = \int_0^\infty \frac{\sin(y)}{y}dy.$$ $\endgroup$ – Hugocito Jul 14 '18 at 2:20
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    $\begingroup$ Awesome thank you so much. I appreciate it. $\endgroup$ – MathIsHard Jul 14 '18 at 2:21
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Notice that $$I=\int^\infty_0\frac{\sin(ax)}x dx=\frac12\int^\infty_{-\infty}\frac{\sin(ax)}x dx=\frac12\Im\underbrace{\text{P}\int^\infty_{-\infty}\frac{e^{iax}}x dx}_{\text{Let}=J}$$

Let $f(z)=\frac{e^{iaz}}z$.

For $a>0$:

Consider a contour $C$, which is an infinitely large semicircle on the upper half plane, centered at the origin. Also add an infinitely small semicircle indent at the origin.

$$\oint_Cf(z)dz=\underbrace{\int_{\text{arc}}f(z)dz}_{\to0}+\underbrace{\int^\infty_{0^+}f(z)dz+\int^{0^-}_{-\infty}f(z)dz}_{=J}+\int_{\text{indent}}f(z)dz$$

By Cauchy’s integral theorem, $$\oint_Cf(z)dz=0$$

Also, $$\int_{\text{indent}}f(z)dz=-\frac12\cdot2\pi i\text{Res}_{z=0}f(z)=-\pi i$$ (the minus sign is due to the clockwise orientation of the indent arc, and the $\frac12$ factor is due to the indent is half of a circle.)

Assembling things, we get $$J=\pi i\implies\color{blue}{ I=\frac{\pi}2}$$

For $a<0$:

Similarly, take an infinitely large semicircle on the lower half plane, centered at origin, with a small indent at the origin.

Here, $$\int_{\text{indent}}f(z)dz=-\frac12\cdot 2\pi i \text{Res}_{z=0}f(z)=-\pi i$$

With similar procedures, by Cauchy’s integral theorem, $$-J-\pi i=0$$

Then, we get $$\color{blue}{I=-\frac{\pi}2}$$

Gluing two case together:

$$\color{red}{\int^\infty_0\frac{\sin(ax)}xdx=\text{sgn}(a)\frac{\pi}2}$$

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  • $\begingroup$ Hey, may you please explain how to change the integral of the semi-circle with radius $\varepsilon$ ($\int_{indent} f(z)dz$) to a full circle with radius $\varepsilon$ in order to apply the residue theorem? (the multiplication by $\frac{-1}{2}$ ). Thanks! $\endgroup$ – dan Jul 6 at 4:05

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