Q:

Evaluating a complex definite integral of $e^{-st^{2}+it}$ from t=0 to t= infinity maybe gamma function??

What would be the value of $\int_{0}^{\infty} e^{-st^{2}+it}dt$ where $i$ is the imaginary unit. I have tried long and hard , tried to use gamma function, substitution , converting to double integral but couldn't find the solution. Does the solution even exist in closed form. If it exists , please help in finding it..

A:

This becomes the Gaussian integral by performing a suitable substitution.
$$\begin{align}
\int_{0}^{\infty} e^{-st^{2}+it}dt
&=\int_{0}^{\infty} e^{-s(t^{2}-\frac{i}st)}dt\\
&=\int_{0}^{\infty} e^{-s((t-\frac{i}{2s})^2+\frac1{4s^2})}dt\\
&=\int_{0}^{\infty} e^{-s(t-\frac{i}{2s})^2-\frac1{4s}}dt\\
&=e^{-\frac1{4s}}\int_{0}^{\infty} e^{-s(t-\frac{i}{2s})^2}dt\\
\end{align}$$
Then applying the substitution $u=\sqrt{s}(t-\frac{i}{2s})\implies du=\sqrt{s}dt$ gives
$$\begin{align}
e^{-\frac1{4s}}\int_{0}^{\infty} e^{-s(t-\frac{i}{2s})^2}dt
&=\frac{1}{e^{\frac1{4s}}\sqrt{s}}\int_{-\frac{i}{2\sqrt{s}}}^{\infty} e^{-u^2}du\\
&=\frac{1}{e^{\frac1{4s}}\sqrt{s}}\int_{-\frac{i}{2\sqrt{s}}}^0 e^{-u^2}du+\frac{1}{e^{\frac1{4s}}\sqrt{s}}\int_0^\infty e^{-u^2}du\\
&=\frac{i\sqrt{\pi}}{2e^{\frac1{4s}}\sqrt{s}}\mathrm{erfi}\left(\frac{1}{2\sqrt{s}}\right)+\frac{\sqrt{\pi}}{2e^{\frac1{4s}}\sqrt{s}}\\
&=\frac{\sqrt{\pi}}{2e^{\frac1{4s}}\sqrt{s}}\left(1+i\mathrm{erfi}\left(\frac{1}{2\sqrt{s}}\right)\right)\\
\end{align}$$

