Hi I have some problem with solving the following definit integral. $$ \int_{-\infty}^{\infty} e^{-\big(\dfrac{t}{\tau}\big)^2-\dfrac{i\omega t}{2}}\textrm{d}t $$ My guess to solve it is by using Fourier transform. I need somehow get write it on the form $$ F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t} $$ And after i get it on this form I can simply look up in a table what $F(\omega)$ is.
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Hint. Alternatively, one may observe that, by parity of the integrand, your integral reduces to $$ \int_{-\infty}^{\infty} e^{-\Big(\dfrac{t}{\tau}\Big)^2-\dfrac{i\omega t}{2}}\textrm{d}t=\int_{-\infty}^{\infty} e^{-\Big(\dfrac{t}{\tau}\Big)^2}\cos \Big(\dfrac{\omega t}{2} \Big)\:\textrm{d}t $$ then one may use the standard result: $$ \int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x=\frac{\sqrt{\pi}}{2} e^{-\large\frac{a^2}{4}}. $$
Olivier Oloa
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May I ask why you didn't include $-i\sin (\dfrac{\omega t}{2})$. Was it because it's an odd function over a even intervall os the integral win sinus will be zero? – Turbotanten Jan 08 '17 at 14:29