I have this integral:
$$ \int_{-\infty}^{a} \frac{1}{(a-x')^{4/3}}\exp{\left(-\frac{{x'}^2}{2b^2}\right)} dx'$$
where $a$ and $b$ are real constants. The integrand is divergent for $x'=a$ and so to evaluate, I attempt to extend the integral to the complex plane:
$$ \int_{C} \frac{1}{(a-z)^{4/3}}\exp{\left(-\frac{{z}^2}{2b^2}\right)} dz$$
where now we have the freedom to choose a contour which will include the $(-\infty,a)$ interval in a simple way. I suppose we have the following options:
-
Infinite Radius Semi-Circle (either Top/Bottom Planes): The contour would wrap around the singularity at $z=(a,0)$, but the difficulty here comes form the fact that at the singularity the gaussian does not tend to zero, and this would make it difficult to compute.
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Infinite Radius Circle Not Including $z=(a,0)$: The contour here would be an infinite radius circle centered at $z=(a,0)$ and a cut on the x-axis to the singularity and another infinitely small circle cutting out the singularity.
I tried working out the two above proposed contours but it does not give useful results. Does anyone have any input or suggestions, or better yet a solution to this integral?
Thank you in advance!
Best Answer
Some may think that it is pointless to post this integral on MSE as it diverges. But, hey, provided that the OP has knowledge in complex analysis, wouldn't this integral be meaningful if a 'principal value' exists?
And yes, it does exist (but not in Cauchy sense).
Define, for $\text{Re } s<1$, $$F(s)=\int_{-\infty}^{\alpha} \frac{1}{(\alpha-x')^s}\exp{\left(-\frac{{x'}^2}{\beta^2}\right)} dx' \stackrel{t=\alpha-x'}{=}\int^\infty_0 t^{-s}\exp\left(-\frac{(t-\alpha)^2}{\beta^2}\right)dt$$
Not every integral is best evaluated by residue theorem. Real methods suit this integral better: $$\begin{align} F(s) &=\int^\infty_0 t^{-s}\exp\left(-\frac{t^2}{\beta^2}\right)\exp\left(\frac{2\alpha t}{\beta^2}\right)\exp\left(-\frac{\alpha^2}{\beta^2}\right)dt \\ &=\exp\left(-\frac{\alpha^2}{\beta^2}\right)\sum^\infty_{n=0}\frac{(2\alpha/\beta^2)^n}{n!}\int^\infty_0 t^{n-s}\exp\left(-\frac{t^2}{\beta^2}\right)dt \\ &\stackrel{v=t^2/\beta^2}{=}\exp\left(-\frac{\alpha^2}{\beta^2}\right)\sum^\infty_{n=0}\frac{(2\alpha/\beta^2)^n}{n!}\frac{\beta^{n-s+1}}2 \int^\infty_0 v^{(n-s-1)/2}e^{-v}dv \\ &=\frac{\beta^{1-s}}{2}\exp\left(-\frac{\alpha^2}{\beta^2}\right) \sum^\infty_{n=0}\frac{(2\alpha/\beta)^n}{n!} \Gamma\left(\frac{n-s+1}2\right) \\ \end{align} $$
From this series form of $F(s)$, it can be easily seen that the series converges whenever $s\not\in\mathbb N$ (thanks to the well-known analytic continuation of $\Gamma$). Therefore, $F(s)$ is immediately analytically continued to $\mathbb C\setminus\mathbb N$.
Responding to your question, $$F\left(\frac43\right)=\frac{\beta^{-1/3}}{2}\exp\left(-\frac{\alpha^2}{\beta^2}\right) \left[\Gamma\left(-\frac16\right)+\sum^\infty_{n=1}\frac{(2\alpha/\beta)^n}{n!} \Gamma\left(\frac n2-\frac16\right)\right]$$
or
$$F\left(\frac43\right)=\frac{\beta^{-1/3}}{2}\exp\left(-\frac{\alpha^2}{\beta^2}\right) \left[-\frac{2\pi}{\Gamma\left(\frac56\right)}+\sum^\infty_{n=1}\frac{(2\alpha/\beta)^n}{n!} \Gamma\left(\frac n2-\frac16\right)\right]$$
in the sense of analytic continuation.
This may be simplified by expressing it in terms of hypergeometric functions, but I am not a big fan of it.