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I have got this integral from a fourier transform:

$$\int_{-\infty}^0 e^{-ikx+x/4}+\int_0^{\infty} e^{-ikx-x/4}dx$$ Apparently the integrals give:

$$=\frac{1}{1/4 -ik}+\frac{1}{1/4 +ik}$$

But how? I'm trying to evaluate the bounds at infinity:

$$\lim_{x\to \infty}\frac{e^{x(-ik-1/4)}}{-ik-1/4}=\infty?$$ but I get infinity when it should converge? I really don't know how to evaluate the bounds.

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  • $\begingroup$ @AnneBauval oops, my bad, editing. $\endgroup$
    – Ivy
    Commented May 11 at 19:37
  • $\begingroup$ What's the limit of $e^{-x/4}\cos kx$ when $x\to\infty$? $\endgroup$
    – Jyrki Lahtonen
    Commented May 11 at 19:50
  • $\begingroup$ @JyrkiLahtonen should be zero, no? $\endgroup$
    – Ivy
    Commented May 11 at 21:32
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    $\begingroup$ Correct. So why did you get $\infty$ on the second line from the bottom? $\endgroup$
    – Jyrki Lahtonen
    Commented May 11 at 21:35
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    $\begingroup$ A general result is that if $x\to\infty$ along the real axis (so not a complex variable) and $\alpha$ is complex number with a negative real part, then $$e^{\alpha x}\to 0.$$ You basically verified this yourself using $$e^{(a+bi)x}=e^{ax}\left(\cos bx+i\sin bx\right).$$ Alternatively $$|e^{(a+bi)x}|=e^{ax},$$ and when this tends to zero, so will the complex limit. $\endgroup$
    – Jyrki Lahtonen
    Commented May 12 at 17:30

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