Working on a Fourier cosine transform and need a solution to the following integral to continue. Very grateful for any responses/ advice. My integration is a little rusty.
Integral ((Sin 2pix Cos piX) / X)
thanks in advance,
-Dave
Working on a Fourier cosine transform and need a solution to the following integral to continue. Very grateful for any responses/ advice. My integration is a little rusty.
Integral ((Sin 2pix Cos piX) / X)
thanks in advance,
-Dave
What are the bounds of the integral? If it is this
$\displaystyle \int_0^{\infty}\frac{\sin(2\pi{x})\cos(\pi{x})}{x} dx$
The answer is $\displaystyle \frac{\pi}{2}$
To see this rewrite your integral as
$\displaystyle \begin{aligned}\int_0^{\infty}\frac{\sin(2\pi{x})\ cos(\pi{x})}{x}dx&=\int_0^{\infty}\sin(2\pi{x})\co s(\pi{x})\int_0^{\infty}e^{-yx}dydx\\
&=\int_0^{\infty}\int_0^{\infty}e^{-yx}\cos(\pi{x})\sin(2\pi{x})dydx
\end{aligned}$
And you may switch the integration order because it is over a rectangular region by Fubini's Theorem.