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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?

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  • $\begingroup$ Use \left( and \right) for larger parentheses. What does "v.p." stand for? $\endgroup$ Mar 20, 2022 at 9:07
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    $\begingroup$ @RodrigodeAzevedo I would guess its the Cauchy principal value $\endgroup$ Mar 20, 2022 at 9:16
  • $\begingroup$ @Egor Wolfram has the ability to compute Fourier transforms. It says you are missing a sign. Screenshot $\endgroup$ Mar 20, 2022 at 9:21
  • $\begingroup$ Oh, I misread your final answer. The minus sign is correct. Sorry! $\endgroup$ Mar 20, 2022 at 9:35
  • $\begingroup$ @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 $\endgroup$
    – Egor
    Mar 20, 2022 at 9:47

1 Answer 1

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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 $\mathcal F$)

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  • $\begingroup$ Everything is clear, but I think that renaming the values makes understanding more difficult. In case, we cannot compare the result sign-by-sign $\endgroup$
    – Egor
    Mar 20, 2022 at 9:51
  • $\begingroup$ Yes, sorry. I just used the notation that I am familliar with to minimise errors. I have changed it! $\endgroup$ Mar 20, 2022 at 11:22

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