# Fourier transform question

assuming that the integral exists

$$I(u)= \int_{-\infty}^{\infty}dxe^{iux}e^{ax}f(x)$$

using the shift properties of Fourier function is that integral equal to

$$I(u)= \frac{F(u+ia)+F(u-ia)}{2}$$

with $$F(u)=\int_{-\infty}^{\infty}dxe^{iux}f(x)$$ ??

or it is just equal to $$I(u)= F(u+ia)$$

what should be the correct solution ? here $f(x)$ real or complex

if $f(x)=f(-x)$ is even then its Fourier Transform must be real but how about in other cases ??

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It seems your question mark key is still stuck. – joriki Dec 11 '12 at 16:01

Those are both wrong; substituting $u-\mathrm ia$ for $u$ in $F(u)$ immediately yields $F(u-\mathrm ia)=I(u)$.