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I came across this integral problem:

$$\hat f(\xi)=\int_{-\infty}^{+\infty} e^{-|x|+xi\xi}dx$$

Now I know how to integrate simple absolute value functions like:

$\int_{-2}^{4}|x-2| dx$, we just find the 'break points', which in this case is $x=2$ so we would integrate $\int_{-2}^2 (x-2) dx + \int_{2}^{4}(2-x)dx$.

But in that first equation we have an imaginary number. How do I deal with it?

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  • $\begingroup$ What's $\xi$? Because if $\xi>0$ then this interval must diverge because for increasingly negative values of $x$ this function increases to infinity. $\endgroup$
    – Gregory Grant
    Feb 12, 2016 at 11:21
  • $\begingroup$ this is the fourier transform of $e^{-|x|}$ $\endgroup$
    – Naz
    Feb 12, 2016 at 11:23
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    $\begingroup$ u forget an $+$ in the exponent, to solve the integral, split the integration range at 0 and do both cases seperatly $\endgroup$
    – tired
    Feb 12, 2016 at 11:24
  • $\begingroup$ Why do I split it at 0? because before that the $-|x|$ will be $x$ and after that it is $-x$? So I have forgotten a plus in the exponent then.... $\endgroup$
    – Naz
    Feb 12, 2016 at 11:27
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    $\begingroup$ $\int_0^\infty e^{- x + i \xi x} dx = \frac{1}{1-i \xi}$ so $\int_{-\infty}^\infty e^{- |x| + i \xi x} dx = 2 Re\left(\frac{1}{1-i \xi}\right)$ $\endgroup$
    – reuns
    Feb 12, 2016 at 11:32

1 Answer 1

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$$\int_{-\infty}^{\infty} e^{-|x| + xi\xi} \ dx = \int_{-\infty}^0 e^{x(1+i\xi)} \ dx + \int_0^{\infty}e^{-x(1-i\xi)} \ dx$$

And then my friend pointed out the following:

$e^{x+ix\xi}=e^xe^{ix\xi}=e^x (\cos(x\xi)+i\sin(x\xi))\leq e^x (1+i)$

Now:

$e^x (1+i)$ - so we can ignore the $(1+i)$ as it is bounded and we only care about $e^x$ which goes to zero as $x \rightarrow -\infty$

therefore:

$$\int_{-\infty}^0 e^{x(1+i\xi)} \ dx + \int_0^{\infty}e^{-x(1-i\xi)} = \frac{1}{1+i\xi} + \frac{1}{1-i\xi}=\frac{2}{1+\xi^2}$$

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  • $\begingroup$ in fact, it can be shown that $\lim_{x \to -\infty} e^{x(1+i\xi)} = 0$, due to sandwich theorem $\endgroup$
    – Naz
    May 18, 2016 at 14:41

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