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.
 A: 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 $x \to - \infty$.
A: 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.
