In my research I encountered the following indefinite integral $$ I=\int x^{n-1/2}e^{x}dx, $$
which looks pretty much like an incomplete Gamma function with the little caveat that in the exponential term I have a positive sign instead of a minus sign. My naive solution was to do the change of variables x=-u:
$$ I=(-1)^{n-1/2}\int u^{n-1/2}e^{-u}du=(-1)^{n-1/2} \Gamma(n+1/2,u), $$
But when plugging my integral in wolfram I get a different result. This is what I get: $$ I=(-x)^{1/2 - n} x^{-1/2 + n} Γ(n+1/2, -x) $$
which differs slightly from what I have. My advisor told me that I should instead consider the analytical continuation of the integral but I really know how to do so, could someone please give me some insight. Thanks in advance.
Your assumption
$$(-u)^{n-\frac{1}{2}}=(-1)^{n-\frac{1}{2}}\ u^{n-\frac{1}{2}}\tag{1}$$
is only valid for $u\ge 0$ and
$$\int x^{n-\frac{1}{2}}\ e^x\,dx=(-x)^{\frac{1}{2}-n}\ x^{n-\frac{1}{2}}\ \Gamma\left(n+\frac{1}{2},-x\right)=-i\ (-1)^n\ \Gamma\left(n+\frac{1}{2},-x\right),\quad x\le 0\tag{2}$$
is equivalent to
$$-(-1)^{n-\frac{1}{2}} \int u^{n-\frac{1}{2}}\ e^{-u}\,du=-i\ (-1)^n\ \Gamma\left(n+\frac{1}{2},u\right),\quad u\ge 0\tag{3}$$
since $x=-u$ and $dx=-du$.