Fourier Cosine Transform and Complex Exponential Solution for Homework Problem

[mhill]

Homework Statement

given 2 functions f and g related by a cosine transform

$$g( \alpha ) = \int_{0}^{\infty}dx f(x)Cos( \alpha x)$$

then if the integral

$$\int_{0}^{\infty}dx f(x)exp(cx)$$

exists for every positive or negative 'c' then should it be equal to

$$\int_{0}^{\infty}dx f(x)exp(cx)= \frac{g(ic)+g(-ic)}{2}$$ ??

Homework Equations

$$g( \alpha ) = \int_{0}^{\infty}dx f(x)Cos( \alpha x)$$

The Attempt at a Solution

where i have used the Euler identity to express the cosine as a linear combination of complex

exponentials.

 

[HallsofIvy]

Yes, that should work. Unfortunately, since you chose not to show us what you did, I can't say where you might have made a mistake.

 

[mhill]

thanks Hallsoftivy.. i think this would be the result since

$$\int_{0}^{\infty}dx f(x)exp(cx)$$ should be real

then i used Euler's formula so $$2exp(cx)Cos(ax)=exp(iax+cx)+exp(-iax+cx)$$

then somehow (of course this all is completely nonrigorous) expanding the exponential into a real and complex part, the contribution to the integral would come from

$$Cos(ax+icx)$$ and $$cos(ax-icx)$$ this kernel is precisely the Kernel of a Fourier cosine transform with complex argument.