DLS
  • DLS
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Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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DLS
  • DLS
\[\Large \int\limits_0^\infty \frac{\sin \alpha \cos \alpha x}{\alpha} d \alpha\] = pi/2 when x=0 to 1 =pi/4 when x=1 =0 when x>1
DLS
  • DLS
I think we can do something with fourier transform on right side and then perform its inverse to show this..not quite sure.
DLS
  • DLS
@ganeshie8 @Jemurray3

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ganeshie8
  • ganeshie8
@Kainui @dan815
Kainui
  • Kainui
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
DLS
  • DLS
This question is from Fourier Transformation in my textbook.
Kainui
  • Kainui
Oh then you should probably try using the convolution theorem if you can
DLS
  • DLS
Can't we take the fourier transform of the RHS to get a function of (say) alpha and then perform the inverse transformation?
Kainui
  • Kainui
what do you mean by right hand side?
Kainui
  • Kainui
Ok I think I see sorta what you're saying, like you wanna do a Fourier transform on this? \[\Large F(x)= \int\limits_0^\infty \frac{\sin \alpha \cos( \alpha x)}{\alpha} d \alpha\]
DLS
  • DLS
\[\Large \int\limits_0^1 \frac{\pi}{4}e^{-i \alpha x} dx\] + so on..and then do the inverse thing to get the same thing as LHS.
DLS
  • DLS
I mean there was a similar example in my book where they did this.^
Kainui
  • Kainui
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.
anonymous
  • anonymous
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 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.
DLS
  • DLS
nice :o you didn't use fourier though :P
anonymous
  • anonymous
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.

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