Evaluating $\int_0^\infty \frac{1}{(k-1)!} (\frac{x}{y})^{k+1} (1-y)^{-x/y} \, dx$ EDIT: I CHANGED THE QUESTION (I HAD THE WRONG BOUNDS!) THE ACTUAL QUESTION WAS FROM 0 TO INFINITY, NOT 0 TO 1!
I'm stuck with evaluating this integral and I need some help!
$$\large\int_0^\infty \frac{1}{(k-1)!} (\frac{x}{y})^{k+1} (1-y)^{-x/y} \, dx$$
My workings are as follows:
$$\int_0^\infty \frac{1}{(k-1)!} \left(\frac{x}{y}\right)^{k+1} (1-y)^{-x/y} \,dx$$
$$\frac{1}{(k-1)!} \int_0^\infty \left(\frac{x}{y}\right)^{k+1} (1-y)^{-x/y} \,dx$$
Let $u = \frac{x}{y}$
$$\frac{1}{(k-1)!} \int_0^\infty  (u)^{k+1} (1-y)^{-u} \,du$$
And now I'm stuck!
 A: As André Nicolas commented, your last integral misses one multiplying $y$ and the bounds are no more correct.
Just working the antiderivative, your last integral should write $$I=\frac{1}{(k-1)!} \int  y~u^{k+1} (1-y)^{-u} \,du=\frac{y}{(k-1)!}\int u^{k+1}e^{-u\log(1-y)}\,du$$ Now, changing variable $u\log(1-y)=z$, this should lead to $$I=\frac{y}{\Big(\log(1-y)\Big)^{k+2}(k-1)!}\int z^{k+1}e^{-z} \,dz$$ and $$J=\int z^{k+1}e^{-z} \,dz=-\Gamma (k+2,z)$$ where appears the incomplete gamma function (as André Nicolas commented).
Edit
The image added to the post shows that  the problem is $$I=\int \frac{(1-y)^{k-1} e^{-\frac{x}{y}} \left(\frac{x}{y}\right)^{k+1}}{(k-1)!}\,dx$$ So, changing $x$ by $u y$ leads to $$I=\int\frac{e^{-u} y u^{k+1} (1-y)^{k-1}}{(k-1)!}\,du$$where appears the same incomplete gamma function. 
A: I don't have 50 reputation to comment, so unfortunately I'll have to leave this as an answer. 
For a lot of these statistics-type problems where you're asked to find the integral of some hairy quantity, a straightforward (and time saving) solution is often to recognize that the integrand is the kernel of a known distribution, then to manipulate the integrand to be a pdf and thus sums to one. 
For instance, is there a constant you can multiply such that the integrand is some known pdf? If so, multiply that inside the integral such that the integral evaluates as one, then multiply by its reciprocal outside the integral. 
The answer should fall out nicely. 
