Question

Show that :

Answer 1

\[\Large \int\limits_0^\infty \frac{\sin \alpha \cos \alpha x}{\alpha} d \alpha\]
= pi/2 when x=0 to 1
=pi/4 when x=1
=0 when x>1

Answer 2

I think we can do something with fourier transform on right side and then perform its inverse to show this..not quite sure.

Answer 3

@ganeshie8 @Jemurray3

Answer 4

@Kainui @dan815

Answer 5

I dunno, I tried taking the derivative with respect to x to see if that helped, but then kinda didn't get anywhere very far

Answer 6

This question is from Fourier Transformation in my textbook.

Answer 7

Oh then you should probably try using the convolution theorem if you can

Answer 8

Can't we take the fourier transform of the RHS to get a function of (say) alpha and then perform the inverse transformation?

Answer 9

what do you mean by right hand side?

Answer 10

Ok I think I see sorta what you're saying, like you wanna do a Fourier transform on this? \[\Large F(x)= \int\limits_0^\infty \frac{\sin \alpha \cos( \alpha x)}{\alpha} d \alpha\]

Answer 11

\[\Large \int\limits_0^1 \frac{\pi}{4}e^{-i \alpha x} dx\] + so on..and then do the inverse thing to get the same thing as LHS.

Answer 12

I mean there was a similar example in my book where they did this.^

Answer 13

I'm not that familiar with doing Fourier transforms, generally when I've used them I just plugged into wolfram alpha since it was faster than looking it up in a table or whatever unfortunately.

Answer 14

It's useful to note that

\[\sin(x)\cos(y) = \frac{1}{2} \left(\sin(x+y) + \sin(x-y)\right)\]
So that integral is equivalent to

\[F(x) =\frac{1}{2} \int_0^\infty \frac{\sin[\alpha(x+1)]}{\alpha} + \frac{\sin[\alpha(1-x)]}{\alpha} d\alpha \equiv \frac{1}{2}\left(F_1(x) + F_2(x)\right)\]

Each one of those integrals can be evaluated without too much of a problem.

\[F_1(x) = \int_0^\infty \frac{\sin[\alpha(1+x)]}{\alpha} d\alpha \]
Changing variables to \[ u=\alpha(1+x)\] we find
\[F_1(x) = \int_0^\infty \frac{\sin(u)}{u} du = \frac{\pi}{2}\]

\(F_2(x)\) is similar. If 0<x<1, the integration will be identical. If x=1, \(F_2\) will be equal to zero. if x > 1, you'll get an extra minus sign, so \(F_2 = -F_1\). You can check that this is exactly what you're looking for. Additionally, if you flip the sign of x, everything will be the same.

Answer 15

nice :o you didn't use fourier though :P

Answer 16

Why would that be necessary? Anytime you have sines and cosines you can transform to exponentials and that's more or less a Fourier transform. Additionally, the entire thing can be viewed as the real part of the Fourier transform

\[F(x) = \frac{1}{2} \int_{-\infty}^\infty \frac{\sin(\alpha)}{\alpha} e^{-i\alpha x} d\alpha \]

but it's important to note that unlike what I did in your previous problem, differentiation with respect to x is no longer allowed. More work is required, and unless you know how to do contour integrals in the complex plane, you'd end up doing some variation of what I already did above.