Using complex contour for calculation integreal sin(x)*exp(ikx)/x

[c0nfig]

Homework Statement

Hi,
i need help with solving the above integral using complex analysis :
\int _{-\infty }^{\infty }\!{\frac {\sin \left( x \right) {{\rm e}^{ikx}}}{x}}{dx}<br />

The Attempt at a Solution

i know that the contour will probably be from -infinity to infinity with indent around the zero point like when solving sin(x)/x integral.
more over, the is actually Fourier transform for sinc , so i think the answer should be a one dim rect function, but i can't see how.
will be grateful for any hints.

 

[HallsofIvy]

I have no idea what you mean by "a one dim rect function" but certainly that can be integrated by using a contour consisting of the line from -R to [math]-\epsilon[/math], a semi-circle of radius [math]\epsilon[/math], the line from [math]\epsilon[/math] to R, a semicircle of radius R, and then taking the limits as R goes to infinity, [math]epsilon[/math] goes to 0.

 

[jackmell]

Homework Statement

Hi,
i need help with solving the above integral using complex analysis :
\int _{-\infty }^{\infty }\!{\frac {\sin \left( x \right) {{\rm e}^{ikx}}}{x}}{dx}<br />

The Attempt at a Solution

i know that the contour will probably be from -infinity to infinity with indent around the zero point like when solving sin(x)/x integral.
more over, the is actually Fourier transform for sinc , so i think the answer should be a one dim rect function, but i can't see how.
will be grateful for any hints.

First, zero is a removable singular point so no need to indent around it. Also, I think you first need to define what k is and I think it diverges if |k|<1. So for now, try and solve it for k>1. In that case, then I think you can do as Hall said above and use Jordan's Lemma or ML inequality maybe to show that the integral over the half-circle contour goes to zero as R goes to infinity and since there are no poles inside the contour, the integral over the real axis is zero too. Try expanding sin(z) in terms of sin(Re^{it}) and combine everything and then show for k>1 and 0<t<pi, the integral (quickly) goes to zero.

 

[c0nfig]

I have no idea what you mean by "a one dim rect function" but certainly that can be integrated by using a contour consisting of the line from -R to [math]-\epsilon[/math], a semi-circle of radius [math]\epsilon[/math], the line from [math]\epsilon[/math] to R, a semicircle of radius R, and then taking the limits as R goes to infinity, [math]epsilon[/math] goes to 0.

thanks for your reply,
just to clarify , rect is just a rectangular function from wiki :
\mathrm{rect}(t) = \sqcap(t) = \begin{cases}<br /> 0 ; \mbox{if } |t| &gt; \frac{1}{2} \\<br /> \frac{1}{2} ; \mbox{if } |t| = \frac{1}{2} \\<br /> 1 ; \mbox{if } |t| &lt; \frac{1}{2}. \\<br />
and it's Fourier transform is simply sin(x)/x (with some constants , and translated if we take the above definition for rectangular function )
now the integral that i asked about is Fourier transform of sin(x)/x so the result should get the rectangular function.

so as HallsofIvy answered : for |k|>1 the integral is zero as rect(k) .
for |k|<1 it shouldn't diverges , but should go to some constant ( actually it is pi )

to do the integral as you said i wrote sin(z) as:( e^iz - e^-iz ) /2i and then multiply it by exp(ikz) . so if i use series now up to first order for the exponents i have : 2iz / 2iz = 1 and the integral is pi.
the problem is it doesn't depend on k because i wrote the series for only to the first order.

HallsofIvy, can you please explain how you got zero there for |k|>1 ?

 

[c0nfig]

First, zero is a removable singular point so no need to indent around it. Also, I think you first need to define what k is and I think it diverges if |k|<1. So for now, try and solve it for k>1. In that case, then I think you can do as Hall said above and use Jordan's Lemma or ML inequality maybe to show that the integral over the half-circle contour goes to zero as R goes to infinity and since there are no poles inside the contour, the integral over the real axis is zero too. Try expanding sin(z) in terms of sin(Re^{it}) and combine everything and then show for k>1 and 0<t<pi, the integral (quickly) goes to zero.

sorry,
jackmell, after expanding i get that the limit for any k is zero.
can you please explain further why you have an answer that depend on k ?

 

[jackmell]

Ok, I was wrong I think about it diverging for |k|<1 because when I plug-in the integral in Mathematica for say k=1/2 and -1/2, it returns pi as you stated above but Mathematica is no guarantee that's correct of course, but usually it is. Also, Mathematica gives zero for say k=2 and -2. Now, I can show it goes to zero for k>1 by just expanding everything in terms of exponents, using Residue Theorem and ML-inequality, but I had problems showing it is zero for k<-1 and I can't figure out how to show it's pi for |k|<1 at least with some quick figuring on the matter. Just needs more work.