On the Fourier transform of $x \mapsto e^{-|x|}$

Question

I am finding Fourier transform quite hard. I am supposed to find the Fourier transform of $$f(x)=e^{-|x|}$$ I tried myself and looked at the solution sheet, but I have some questions.

1. How come $ \int_{-\infty}^{\infty} e^{-|x|}\sin(wx)dx = 0$?

2. How come they change the integration $\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-|x|}\cos(wx)dx$ = $\sqrt{\frac{2}{\pi}}\int_{0}^{\infty} e^{-|x|}\cos(wx)dx$?

Answer 1

In answer to your question 1, the integrand is odd so the integral is zero.

For question 2, the LHS lower limit was $-\infty$ not $0$. The integrand is even so the integral from $-\infty$ to $+\infty$ is $2\times$ the same integral from $0$ to $\infty$.

Answer 2

howcome $\int_{-\infty}^{\infty}e^{-|x|}\sin(\omega x)\,dx=0$?

notice that if: $$f(x)=e^{-|x|}\sin(\omega x)$$ then: $$f(-x)=-f(x)$$ i.e. it is an odd function. This means that: $$\int_{-\infty}^0f(x)\,dx=-\int_0^\infty f(x)\,dx\therefore \int_{-\infty}^0f(x)\,dx+\int_0^\infty f(x)\,dx=0$$

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howcome they change the integration $\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-|x|}\cos(wx)dx$ = $\sqrt{\frac{2}{\pi}}\int_{0}^{\infty} e^{-|x|}\cos(wx)dx$

Notice that the integrand is this time an even function i.e. $g(-x)=g(x)$ (also note you made a typo in the lower limit of LHS). This means: $$\int_{-\infty}^0g(x)\,dx=\int_0^\infty g(x)\,dx\therefore \int_{-\infty}^\infty g(x)\,dx=2\int_0^\infty g(x)\,dx$$ Now you just have to do some rearranging: $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-|x|}\cos(\omega x)\,dx=2\frac{1}{\sqrt{2\pi}}\int_0^\infty e^{-|x|}\cos(\omega x)\,dx=\sqrt{\frac 2 \pi}\int_0^\infty e^{-|x|}\cos(\omega x)\,dx$$ and you obtain the next line

Answer 3

Do it like this $$F(w)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-|x|} e^{-iwx} dx=\frac{1}{\sqrt{2\pi}} \left (\int_{-\infty}^{0}e^{x} e^{-iwx} dx +\int_{0}^{\infty}e^{-x} e^{-iwx} dx \right)$$ In the first let $x=-y$, then $$F(w)=\frac{1}{\sqrt{2\pi}} \left(\int_{0}^{\infty} e^{-(1-iw)y} dy + \int_{0}^{\infty} e^{-(1+iw)x} dx \right)$$

$$\implies F(w)=\frac{1}{\sqrt{2\pi}}\left(\frac{1}{1-iw}+\frac{1}{1+iw}\right)$$ $$ =\sqrt{\frac{2}{\pi}}\frac{1}{1+w^2}$$