Calculation of $\int_0^\infty \frac{x^{1-a}}{x+c} dx $ using Gamma function I am supposed to calculate some integrals using Gamma functions.
The simplest of the integrals is
$$ I(c)=\int_0^\infty \frac{x^{1-a}}{x+c} dx  $$
and the two relations I am supposed to use are
$$ \frac{\Gamma(b+1)}{a^{b+1}}=\int_0^\infty t^be^{-at}dt $$
and Euler's reflection formula,
$$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)} $$
Try as I might, I have not managed to correlate the formulas to the integral in any significant way (in large part due to that pesky exponential I can't seem to shake). The supposed result of the above integral is $$I(c)=-\frac{\pi}{\sin(\pi a)}c^{1-a}$$
I definitely don't want to outsource my responsibilities (and once I understand how to go about this I'm confident I'll work out the more complicated integrals myself), but I've hit a wall here. I'd appreciate either a solution or a nudge in the right direction.
 A: This is an outline of a solution, and I hope it's detailed enough for you to fill in the gaps.
I assume you have attempted to use the two relations by substituting the first relation into the second one, which would get you something like
$$
\frac{\pi}{\sin(\pi b)} = \Gamma(b+1)\Gamma(-b) = a \int_0^\infty \int_0^\infty s^{-b-1} t^{b} e^{-a(s+t)} ds dt
$$
The hope is to somehow transform the right hand side into some form that is proportional to $I(c)$. This is in fact the correct direction to work towards.
You said that you can't get rid of the pesky exponential factor, but you can probably also guess that the key is to come up with some substitution where one of the new variables is $u = s+t$, so that the integral with respect to $u$ will get rid of the exponential factor. The substitution that will work is in fact $s = u v$, $t = u(1-v)$. Note that the $u$-dependent part of $s^{-b-1} t^{b}$ will simplify to $u^{-1}$. Combined with the Jacobian factor $u$, the only remaining $u$ dependence in the integrand is $e^{-au}$. Thus integrating with respect to $u$ will get rid of the exponential factor.
The remaining integral should be of the form
$$
\int_0^1 v^{-b-1} (1-v)^{b} dv\ .
$$
A substitution along the lines of $x = \frac{v^2}{1-v^2}$ will bring the integral closer to the form in which $I(c)$ is originally written down.
Note that the relation you are trying to prove can be thought of as a special case of the relation
$$
\mathrm{B}(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\ ,
$$
that is, the case when $x+y = 1$, and the integral $I(c)$ can be seen as a multiple of one of the integral representations of the beta function (see equation (22)),
$$
\mathrm{B}(2-a, a-1) = \int_0^\infty \frac{x^{1-a}}{1+x}dx\ .
$$
The proof outline I sketched above is a specialization of a standard proof the relation between the two functions. In fact the proof of the general case is more natural and suggests the first substitution used above.
