- #1
J6204
- 56
- 2
Homework Statement
Considering the function $$f(x) = e^{-x}, x>0$$ and $$f(-x) = f(x)$$. I am trying to find the Fourier integral representation of f(x).
Homework Equations
$$f(x) = \int_0^\infty \left( A(\alpha)\cos\alpha x +B(\alpha) \sin\alpha x\right) d\alpha$$
$$A(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\cos\alpha u du$$
$$B(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\sin\alpha u du$$
The Attempt at a Solution
$$A(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\cos\alpha u du = \frac{1}{\pi} \int_0^\infty e^{-u}\cos \alpha u du$$**Notice here how I used 0 and iinfinity as my bounds, is this correct?**
Below is the calculation of this integral, I will save myself sometime and just post the answer I got.
$$A(\alpha) = \frac{1}{\alpha ^2 +1}$$
Calculting $$B(\alpha),$$
$$B(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\sin\alpha u du = \frac{1}{\pi} \int_0^\infty e^{-u}\sin \alpha u du$$
**Notice again here how I used 0 and infinity as my bounds, is this correct?**
Below is the calculation of this integral, I will save myself sometime and just post the answer I got.
$$B(\alpha) = \frac{\alpha}{\alpha^2 +1}$$