# Check the Fourier transform for a function: $F[f]=?,f(x)=xe^{-\alpha|x|}$

We have a function to transform (get $$F[f(x)]$$):

$$f(x)=xe^{-\alpha|x|}$$

Using this formula (v.p. meaning):

$$F[f(y)]=v.p.\frac1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{f(t)e^{-ity}dt}$$

In my school, I went to the board and solved this. But now I tried to solve it again, and got other answer (other sign).

My way:

$$F[f(y)]= \frac1{\sqrt{2\pi}}\left(\int_{-\infty}^{0}{(f(t)e^{-ity}dt)}+\int_{0}^{+\infty}{(f(t)e^{-ity}dt)}\right)= \frac1{\sqrt{2\pi}}(\int_{-\infty}^{0}{(te^{\alpha t}e^{-ity}dt)}+\int_{0}^{+\infty}{(te^{-\alpha t}e^{-ity}dt)})= \frac1{\sqrt{2\pi}}(\int_{-\infty}^{0}{(te^{\alpha t-ity}dt)}+\int_{0}^{+\infty}{(te^{-\alpha t-ity}dt)})$$

Replace with: ($$\alpha t-ity=t\beta$$) and ($$-\alpha t-ity=t\gamma$$) and continue:

$$F[f(y)]= \frac1{\sqrt{2\pi}}(\int_{-\infty}^{0}{(te^{t\beta}dt)}+\int_{0}^{+\infty}{(te^{t\gamma}dt)})$$

And I just computed this two integrals using Wolframalpha:

$$F[f(y)]= \frac1{\sqrt{2\pi}}(-\frac1{\beta^2}+\frac1{\gamma^2})$$

Now we should just recover replaced, simplify, and result will be like:

$$F[f(y)]=-\frac{4\alpha i y}{\sqrt{2\pi}(\alpha^2+y^2)^2}$$

Is it correct?

• Use \left( and \right) for larger parentheses. What does "v.p." stand for? Mar 20, 2022 at 9:07
• @RodrigodeAzevedo I would guess its the Cauchy principal value Mar 20, 2022 at 9:16
• @Egor Wolfram has the ability to compute Fourier transforms. It says you are missing a sign. Screenshot Mar 20, 2022 at 9:21
• Oh, I misread your final answer. The minus sign is correct. Sorry! Mar 20, 2022 at 9:35
• @CalvinKhor Yes, thank you, it was my fault that I was not giving v.p. meaning. And yes, you have presented true meaning of it
– Egor
Mar 20, 2022 at 9:47

Result is correct (and there is no need for the principal value as the integrand is continuous and integrable). Wolfram can check 1D Fourier transforms (pic). $$\renewcommand{\xi}y$$
Another way to get the result by hand. Perhaps you know the result ($$\mathcal Ff(\xi):=\int_{-\infty}^\infty f(x)e^{-ix\xi}dx)$$ $$F_1(\xi):= \mathcal F(\exp(-|x|))(\xi)=\frac2{\xi^2+1}$$ Scaling: $$F_\alpha(\xi):=\mathcal F(\exp(-\alpha|x|)(\xi)=\frac1\alpha F_1\left(\frac\xi \alpha\right) = \frac{2\alpha}{\xi^2+\alpha^2}$$ Since $$\mathcal F (-ixf) = \int_{-\infty}^\infty f(x) (\partial_\xi e^{-ix\xi}) dx = \partial_\xi \mathcal F(f)(\xi)$$ we have $$\mathcal F(x\exp(-\alpha|x|))(\xi)= i\partial_\xi F_\alpha(\xi) = i \cdot 2k \cdot \frac1{(\xi^2+\alpha^2)^2}\cdot (-1) \cdot 2\xi = \frac{-4i\alpha\xi}{(\xi^2+\alpha^2)^2}$$
which is your answer (up to convention for defining $$\mathcal F$$)