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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

1 Answer 1

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

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