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.

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.
