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I wonder if anyone could confirm my solution to this integral? $$ \int_0^\infty x e^{-iax^2/2}dx = -\frac{i}{a}, \quad \Im[a]<0 \tag{1} $$ I've included the answer that I got from plugging in the integral to mathematica.

To solve this integral I think that I have to assume that $a$ has a small imaginary part i.e. $a\rightarrow a-i\varepsilon$ where $\varepsilon>0$, and then use Cauchy's integral theorem. $$ \begin{align} \oint_\Gamma z e^{-iaz^2/2}dz&=0\\&=\int_0^R x e^{-i(a-i\varepsilon)x^2/2}dx +iR^2\int_0^{\pi/4} e^{i2\phi}e^{-i(a-i\varepsilon)R^2\text{exp}(i2\phi)/2}d\phi +i\int_R^0 \rho e^{(a-i\varepsilon)\rho^2/2}d\rho \end{align} $$ I think I then should argue that the second integral vanishes as $R\rightarrow\infty$ because the exponential doesn't change sign within the integration region. If I then let $\varepsilon\rightarrow0^+$, I get $$ \int_0^\infty x e^{-iax^2/2}dx=-i\int_\infty^0 \rho e^{a\rho^2/2}d\rho=-\frac{i}{a} $$ What I'm most concerned about is if I can let $\varepsilon\rightarrow0$ before I've completed the integration? Because right it feels like I could just have forgotten the $i\varepsilon$ prescription because It seems to me it dosen't add anything to my problem.

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$$\int_{0}^\infty xe^{-\frac{iax^2}{2}}dx$$ Let $u=\frac{x^2}{2}$ and $du=xdx$

So $$\int_{0}^\infty xe^{-\frac{iax^2}{2}}dx=\int_{0}^\infty e^{-iau}du=\frac{i}{a}(e^{-iau}|_{0}^\infty)$$

If we replace the expression with $\lim_{b\to 0} e^{-(i+b)au}$, you can see that evaluation at infinity is equal to zero.

Thus you are correct and the answer is $-\frac{i}{a}$.

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  • $\begingroup$ That's a nice and short solution to the problem. I like it. Much easier to understand than perhaps this crosspost $\endgroup$ – Turbotanten Jan 24 '18 at 15:06

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