Proving & Solving Integrals with Multiplication Theorem

[matematikuvol]

Homework Statement

Prove

\sqrt{\frac{2}{\pi}}\int^{\infty}_0x^{-\frac{1}{2}}\cos (xt)dx=t^{-\frac{1}{2}}

and use that to solve

\int^{\infty}_0\cos y^2dy

Is this good way to try to prove?

Homework Equations

The Attempt at a Solution

Homework Statement

Multiplicate both sides with \cos x'tdt and integrate from zero to \infty

\sqrt{\frac{2}{\pi}}\int^{\infty}_0dt\cos (x't)\int^{\infty}_0x^{-\frac{1}{2}}\cos (xt)dx=\int^{\infty}_0dt\cos (x't)t^{-\frac{1}{2}}=\sqrt{\frac{2}{\pi}}\int^{\infty}_0dxx^{-\frac{1}{2}}\int^{\infty}_0dt\cos (x't)\cos (xt)dx

 

[matematikuvol]

Suppose that we know

\sqrt{\frac{2}{\pi}}\int^{\infty}_0\cos(xt)x^{-\frac{1}{2}}dx=t^{-\frac{1}{2}}

without proving. How to calculate then

\int^{\infty}_0\cos x^2dx

 

[crimsonidol]

you need to use residue calculus.
if you can go to library look at hildebrand advanced calculus for applications, under the intended contours you will see how to use cauchy's principle and then you'll get gamma functions.

 

[matematikuvol]

For which part of problem. This is problem from Arfken, Weber.

 

[crimsonidol]

you can do the proof and also find part b when you understand the first part I assume.look hildebrand page 561 to be exact

 

[matematikuvol]

\int^{\infty}_0\frac{\cos x}{x^{1-m}}dx=\Gamma(m)\cos (\frac{m\pi}{2})

\int^{\infty}_0\frac{\cos x}{x^{1-\frac{1}{2}}}dx=\Gamma(\frac{1}{2})\cos (\frac{\frac{1}{2}\pi}{2})=\sqrt{\pi}\frac{\sqrt{2}}{2}=\sqrt{\frac{\pi}{2}}

I don't see solution :(

 

[crimsonidol]

you need to look at contour integration and use xt instead of x there. By using residue and appropriate contour you'll be able to find t^-1/2

 

[matematikuvol]

Ok. Thanks. And what then. When I prove first part, how can I calculate integral \int^{\infty}_0\cos x^2dx?

 

[crimsonidol]

https://www.physicsforums.com/showthread.php?t=139465

 

[dextercioby]

Take t=1 in what you have proven already then do a simple substitution to get rid of the sqrt.

 

[matematikuvol]

Thanks a lot! :)