4

I am able to calculate $\int_{0}^{\infty} f(x) \cos(x)\, \mathrm{d}x$ for $f(x)$ being even by taking the real part of Complex Fourier transform (at $\omega = 1$). The two-sided sine transform is $0$, as $f(x)$ is even.

Is there a way to calculate the one-sided integral $\int_{0}^{\infty} f(x) \sin(x) \,\mathrm{d}x$ from the complex fourier transform $\int_{-\infty}^{\infty} f(x) e^{i \omega x} \,\mathrm{d}x$ when $f(x)$ is even?

Brian61354270
  • 2,170
Srini
  • 823
  • To add more context about what Itried from my side: I converted the integral $\int_{0}^{\infty} f(x) sin(x) dx$ to $\int_{-\infty}^{\infty}f(x) \theta(x) sin(x) dx$, where $\theta(x)$ is heaviside function or step function. Then I treat that as $Im[FT(f(x) \theta(x))]$, where FT is fourier transform and Im is imaginary part. Then I expressed the FT as a convolution of $FT(f(x)) \star FT(\theta(x))$, where $\star$ is convolution. Sure, that is one solution I guess, but looks way too complicated – Srini Nov 10 '21 at 20:41

0 Answers0