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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..

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  • $\begingroup$ Try completing the square on $-st^2+it$. $\endgroup$ – dxdydz Apr 15 at 16:58
  • $\begingroup$ WA says this here $$\text{ConditionalExpression}\left[\frac{\sqrt{\pi } e^{\left.-\frac{1}{4}\right/s} \left(1+i \text{erfi}\left(\frac{1}{2 \sqrt{s}}\right)\right)}{2 \sqrt{s}},\Re(s)>0\right]$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 15 at 17:06
  • $\begingroup$ Sorry , I didn't get you.. how would I do completing the square here?? @dxdydz $\endgroup$ – Jeevesh Juneja Apr 15 at 17:07
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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 $$\begin{align} e^{-\frac1{4s}}\int_{0}^{\infty} e^{-s(t-\frac{i}{2s})^2}dt &=\frac{1}{e^{\frac1{4s}}\sqrt{s}}\int_{-\frac{i}{2\sqrt{s}}}^{\infty} e^{-u^2}du\\ &=\frac{1}{e^{\frac1{4s}}\sqrt{s}}\int_{-\frac{i}{2\sqrt{s}}}^0 e^{-u^2}du+\frac{1}{e^{\frac1{4s}}\sqrt{s}}\int_0^\infty e^{-u^2}du\\ &=\frac{i\sqrt{\pi}}{2e^{\frac1{4s}}\sqrt{s}}\mathrm{erfi}\left(\frac{1}{2\sqrt{s}}\right)+\frac{\sqrt{\pi}}{2e^{\frac1{4s}}\sqrt{s}}\\ &=\frac{\sqrt{\pi}}{2e^{\frac1{4s}}\sqrt{s}}\left(1+i\mathrm{erfi}\left(\frac{1}{2\sqrt{s}}\right)\right)\\ \end{align}$$

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  • $\begingroup$ Which result have you used sir while obtaining the limits of second substitution and subsequently breaking the integration into 2 part? $\endgroup$ – Jeevesh Juneja Apr 15 at 17:45
  • $\begingroup$ I only applied one substitution, so you'll have to explain which part. To split the integral I just used that $\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^b f(x)dx$ for all $c$. $\endgroup$ – Peter Foreman Apr 15 at 17:57
  • $\begingroup$ I meant $u$ varies from $\frac{-i}{2\sqrt{s}}$ to $\infty-\frac{i}{2\sqrt{s}}$ then how does it break into $\frac{-i}{2\sqrt{s}}$ to $0+$other term?? Shouldn't $\frac{-i}{2\sqrt{s}}$ stay throughout the integral?? @Peter Foreman $\endgroup$ – Jeevesh Juneja Apr 15 at 18:05
  • $\begingroup$ I never wrote $\infty-\frac{i}{2\sqrt{s}}$ I just wrote $\infty$ because $\mathrm{erf}(\infty + a)=\mathrm{erf}(\infty)$ for all $a$ $\endgroup$ – Peter Foreman Apr 15 at 18:12
  • $\begingroup$ At first it is an integral whose input is along a line parallel to the real axis at a distance of $\frac{1}{2\sqrt{s}}$ below it . Then you transformed it into an integral whose input is along the imaginary axis from $\frac{-i}{2\sqrt{s}}$ to $0$ + an integral whose input is along real axis from $0$ to $\infty$ . Which result enables you to do so? Do you get my point now?? $\endgroup$ – Jeevesh Juneja Apr 15 at 18:19

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