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So I'm taking a complex analysis course on EdX, and this is the first problem on the first problem set. It's far overdue, so I'm not looking to submit it. I just want to know what are the inner workings of how one gets from the question to the answer.

The integral is: $$K(\lambda) = \int_{-\infty}^{\infty}e^{-\lambda t^2}\frac{dt}{t-ib}$$

The equation that it apparently satisfies is (and I don't know this to start with, this is just from clicking on Show Answer; there are blanks in place of the $'$, $b^2$, and $\displaystyle \frac {b}{\sqrt{\lambda}}$):

$$K' - b^2K=-i\sqrt{\pi}\frac{b}{\sqrt{\lambda}}$$

This is a differentiation w.r.t. parameter problem, so $\displaystyle K'\equiv\frac{dK}{d\lambda}$.

I've tried expanding the $\displaystyle \frac{1}{t-ib}$ term, but that doesn't seem to lead anywhere. I've tried differentiating the $e^{-\lambda t^2}$ term w.r.t. parameter, but that doesn't make the expression look like anything nice either.

I don't know how to go about this, and this is quite literally the first problem of a course in which I can follow the lectures.

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I'll assume the instructor has shown why $\int_{-\infty}^\infty e^{-\lambda t^2}dt=\sqrt{\pi/\lambda}$ (if not, look up "Gaussian integral"). We have

$$ K'(\lambda)-b^2K(\lambda)=-\int_{-\infty}^\infty e^{-\lambda t^2}\left(\frac{t^2+b^2}{t-ib}\right)dt=-\int_{-\infty}^\infty e^{-\lambda t^2}(t+ib)dt. $$

The $t$ term integrates to zero since $t\mapsto t\,e^{-\lambda t^2}$ is an odd function, and for any integrable $f$, we have $\int_{-\infty}^\infty f(t)dt=\int_{-\infty}^\infty f(-t)dt$ after changing variables.

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  • $\begingroup$ I should probably edit this, but I don't know the end-point differential equation before-hand. I'm supposed to reach that end point. So I don't a priori know that -b^2 coefficient. Thank you, though. This gives me a place to attempt again. $\endgroup$ – Chronum Jun 15 '17 at 14:38
  • $\begingroup$ Fair enough. I wish you luck! $\endgroup$ – user254433 Jun 15 '17 at 14:43
  • $\begingroup$ On second thoughts, this appears to be more of an intuition problem (I should be able to intuitively see that the t^2+b^2 has something to do with the denominator), and your answer appears to be the way to go. $\endgroup$ – Chronum Jun 15 '17 at 14:45
  • $\begingroup$ Here's some extra intuition in case it helps. We know how to evaluate $\int e^{-t^2}dt$ and $\int t e^{-t^2}dt$, but not $\int \frac{1}{t+i}e^{-t^2}dt$, so it's natural to try to "increase the polynomial order" of the integrand. Differentiating with respect to $\lambda$ gives $\frac{t^2}{t-ib}$, so we're nearly there, but we want the denominator to cancel with the numerator. To make it cancel, add and subtract: $\frac{(t^2-ibt)+ibt}{t-ib}=t+\frac{ibt}{t-ib}$. Add and subtract again: $\frac{ibt-(ib)^2+(ib)^2}{t-ib}=ib-\frac{b^2}{t-ib}$. But the second term corresponds to $-b^2K(\lambda)$. $\endgroup$ – user254433 Jun 15 '17 at 14:58
  • $\begingroup$ Thank you so much for clarifying all of this. I'm still getting used to the way of thinking about all of this. It's been a long time since I've attempted to learn something this far out of my league. Can't say it feels bad, but then again, being stuck doesn't feel good. $\endgroup$ – Chronum Jun 15 '17 at 15:04

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