# Contour integral.

Consider the function $y(x)$ defined by $$y(x)=e^{x^2}\int_{C_1'}\frac{e^{-u^2}}{(u-x)^{n+1}}du$$where $C_1'$ is as shown

The Author makes following claims regarding the behavior of $y(x)$ in the limit of large $x$ (It is assumed that $n>-\frac{1}{2}$, but not integral).

1) As $x\rightarrow+\infty$, the whole path of integration $C_1'$ moves to infinity, and the integral in the above expression tends to zero as $e^{-x^2}$.

2) As $x\rightarrow-\infty$, however, the path of integration extends along the whole of real axis, and the integral in the expression does not tend $\boldsymbol{exponentially}$ to zero, so the function $y(x)$ becomes infinite essentially as $e^{x^2}$.

In regard to the second claim, I can see that the integrals on the parts of the contour above and below the real axis will not cancel since $n+1$ is not integral. I understand these estimates are correct but have not been able to exactly see how. Any indication in the right direction would be very useful.

Thanks.

• You might consider translating the variable of integration, $u\to x+z$, so that the (weighted) integrand becomes $\ e^{-z(2x+z)}z^{-1-n}dz\$ and the translated contour is a tight loop around the positive real axis. – will Dec 30 '15 at 1:22
• @will I, in fact, did try that. But it remained unclear from the transformed integral as well. In particular, is it possible to apply a steepest descent approach to this integral? – vnd Dec 30 '15 at 2:01
• Taking the steepest descent would provide an upper bound. But deforming the contour is a good idea. Because the integrand vanishes as $u\to+\infty$, you could replace $C_1'$ with a loop around $x.$ The tighter the loop, the larger $(u-x)^{-n-1}$ becomes, where a loose loop, like $|u-x| = 1,$ has an exponentially increasing $e^{x^2-u^2}$. – will Dec 31 '15 at 2:32

I assume the circle part is the circle of radius $1$ around $x$. Then the path is separated in three parts, two halflines on the real axis, and the circle.

In the circle part, the integrand is bounded by $e^{-(|x|-1)^2}$ and its length is $2\pi$, so the integral is a $O(e^{-x^2+2x})$.

Then your author then seems to claim that the integral on the halfline doesn't converge to $0$. I disagree with that, $\int_\Bbb R e^{-u^2} du$ is finite, and $\frac 1{(x-u)^{n+1}} \le 1$ for $u \ge x+1$, so you can apply the dominated convergence theorem to show that $\lim_{x \to - \infty} \int_{x+1}^\infty \frac {e^{-u^2}}{(u-x)^{n+1}} du = 0$

Since both integrals tend to $0$ as $x \to - \infty$, we get that $g(x)$ is a $o(e^{x^2})$.

More precisely, the halfline integral should be a $\Theta(|x|^{-n-1})$ as $x \to - \infty$, this makes $y(x)$ a $\Theta(e^{x^2}|x|^{-n-1})$ as $x \to - \infty$.

• Thanks for pointing that the integral over the half lines does indeed tend to 0. I made the mistake while typing the Authors' claim, so have edited it accordingly. – vnd Jan 2 '16 at 21:27

In my point of view the better way to understand behavior of this function is to make changes of variables $v=u+x$.

Thus $y(x)$ has the following form $$y(x)=e^{x^2}\int_{C''}\frac{{e^{-(v+x)^2}}}{v^{n+1}}dv$$. Since $n+1$ is not integer one can represent it in terms of usual (not contour) integral $$y(x)=(1-e^{-2i\pi n})e^{x^2}\int_{C''}\frac{{e^{-(v+x)^2}}}{v^{n+1}}dv$$.

I will omit factor $(1-e^{-2i\pi n})$.

This integral can be expressed via confluent confluent hypergeometric function.

$$y(x)=\frac{\Gamma(-n/2)}{2} F(-n/2,1/2,x^2)-x\Gamma(1/2-n/2)F(1/2-n/2,3/2,x^2)$$, where $F$ is a confluent hypergeometric function. For x>0 this expression has the following form $$y(x)=2^n \Gamma(-n)U(-n/2,1/2,x^2)$$, where U is another solution of confluent hypergeometric equation.

• Should it not be $u=v+x$ rather than $v=u+x$ in the first line of your answer? Anyway that is a trivial point. I will try to follow your argument later. Thanks. – vnd Dec 30 '15 at 6:24