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I'm trying to compute the Fourier Transform of $e^{-|x|}$ however I receive a different answer than wolfram alpha. Wolfram Alpha gets $\sqrt{\frac{2}{\pi}}\frac{1}{1+\varepsilon^2}$ and I get $\frac{2}{1+\varepsilon^2}$. I'll illustrate my working below:

$$F_{x \rightarrow t}(e^{-|x|})=\int_{-\infty}^{\infty}e^{i\varepsilon x} \cdot e^{-|x|} dx=\int_0^{\infty}e^{x(i\varepsilon -1)}dx+\int_{-\infty}^0 e^{x(i\varepsilon +1)} dx $$ And since $$e^{i\varepsilon x}e^{-x}\leq\left | e^{i \varepsilon x}\right| \cdot \left | e^{-x}\right |=\left | e^{-x}\right | \Rightarrow lim_{a \rightarrow \infty} e^{a(i\varepsilon -1)}=0$$ And similarly for $-a$. We have the integrals converging to $$\frac{-1}{i\varepsilon -1}+\frac{1}{i\varepsilon +1}=\frac{2}{1+\varepsilon^2}$$ Where does the factor of $\sqrt{\frac{2}{\pi}}$ come from?

copper.hat
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George1811
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  • That is because there is a factor $1/\sqrt{2\pi}$ when computing Fourier transform. – Eclipse Sun Mar 11 '15 at 14:56
  • This must be a different definition than the one I was taught. My inverse fourier has a factor of $\frac{1}{2\pi}$ – George1811 Mar 11 '15 at 14:57
  • It is just a matter of convention. – Eclipse Sun Mar 11 '15 at 14:59
  • Ah thanks very much, this was making me extremely confused – George1811 Mar 11 '15 at 14:59
  • There are different definitions, some folks use ${1 \over \sqrt{2 \pi}}$ to get some sort of symmetry between the transform and its inverse. – copper.hat Mar 11 '15 at 14:59
  • You're using a different definition of the FT from that of WA. There is no standard definition. Some write the FT with a positive exponent, while others use a negative one. Some use a factor of $1$, while others use a factor of $\frac{1}{2\pi}$, while others still use a factor of $\frac{1}{\sqrt{2\pi}}$ – Mark Viola Mar 11 '15 at 15:06
  • If there is no distinction between negative and positive exponents does this mean a result shown with a positive exponent holds with a negative exponent? – George1811 Mar 11 '15 at 15:14

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Wolfram Alpha defines the Fourier transform of an integrable function as $$ \hat{f}(\xi ) =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-i\xi x} \, dx, $$ while the inverse Fourier transform is taken to be $$ \check{f}(\xi ) =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{i\xi x} \, dx. $$ If you check your solution and multiply it by the factor $1/\sqrt{2\pi}$ you will obtain Wolfram Alpha's one. The reason why you don't get the same result is because you are using the following definition for the F. transform: $$ \hat{f}(\xi ) = \int_{-\infty}^{\infty} f(x) e^{-i\xi x} \, dx, $$ but then you have to move the $1/\sqrt{2\pi}$ factor into the inverse transform, i.e. $$ \check{f}(\xi ) =\frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) e^{i\xi x} \, dx. $$ So, whatever definition you take, be careful because it will condition your inverse F. transform.

Mark Viola
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Alberto Debernardi
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    There is no "usually." Some write the FT with a positive exponent, while others use a negative one. Some use a factor of $1$, while others use a factor of $\frac{1}{2\pi}$, while others still use a factor of $\frac{1}{\sqrt{2\pi}}$ – Mark Viola Mar 11 '15 at 15:08
  • Edited, thanks! – Alberto Debernardi Mar 11 '15 at 15:10
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    That looks great! +1 One more thing, you state that the OP used both definitions. I believe that he used one and WA uses the other. – Mark Viola Mar 11 '15 at 15:24