Fourier Integral for Cosine Function with Infinite Limits

  • Thread starter Thread starter kasse
  • Start date Start date
  • Tags Tags
    Infinite Integral
kasse
Messages
383
Reaction score
1
Working with Fourier integrals, I need to find the integral

\int cos (wv) dv between -\infty and \infty. Is it possible to find this integral?

What I get is sin(wv)/(pi*w) with the infinite limits for v.
 
Physics news on Phys.org
Where did the pi come from? In the sense of a ordinary integral it makes no sense, it's divergent. In the sense of a Fourier integral it can make some sense, if you are also integrating over w. But I think you need to give us the complete context.
 
If you are working with Fourier series, why are you integrating from negative infinity to infinity? The integral should be over one period.

If you are working with the Fourier transform, why are you integrating cosine? Write everything in terms of exponentials: cos(wv)= (ewv+ e-wv)/2.

As Dick said, please give us the entire problem.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top