We start with the following integral
$$I:=\int_0^\infty\mathrm{d}\alpha\,(a+\alpha)^{-\lambda}$$
which is easily evaluated to ${\frac {{a}^{-\lambda+1}}{\lambda-1}}$ by direct integration over $\alpha$ assuming that $a>0,\lambda>1$. This is not what I want to do. In my real-world case the situation is more complicated: I have multiple integrals over $\alpha_1,\ldots,\alpha_n$ from $0$ to $\infty$, and the integrand is $(f(\alpha_1,\ldots,\alpha_n))^{-\lambda}$ where $f$ is a polynom of degree $n$ of $\alpha_1,\ldots,\alpha_n$. Though the problem I stumbled upon occurs in the simplified integral $I$:
Given the Mellin-Barnes representation
$$ \frac{1}{(a+b)^\lambda} = \frac{1}{\Gamma(\lambda)}\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \mathrm{d}z\,\Gamma(-z)\Gamma(\lambda+z)\frac{a^z}{b^{\lambda+z}}$$
(The contour is closed such that poles of $\Gamma(-z)$ are separated from those of $\Gamma(\lambda+z)$; the residues will give a geometric series that leads to the representation.)
I can apply it to the integrand of $I$:
$$I=\int_0^\infty\mathrm{d}\alpha\,(a+\alpha)^{-\lambda}=\frac{1}{\Gamma(\lambda)}\frac{1}{2\pi i}\int_0^\infty\mathrm{d}\alpha \int_{c-i\infty}^{c+i\infty} \mathrm{d}z\,\Gamma(-z)\Gamma(\lambda+z)\frac{a^z}{\alpha^{\lambda+z}}$$
My plan was now to integrate over $\alpha$, since the sum in the denominator was converted to a product, and the $\alpha$-integration should be trivial. The problem here is that the integration is from $0$ to $\infty$ and the integration over $\alpha$ is not possible here.
Can someone point me to the source of the problem in this situation?
For those familiar with physics: $\alpha$ is a Schwinger parameter coming from parametrizing a Feynman loop integral (https://en.wikipedia.org/wiki/Schwinger_parametrization). One usually performs Feynman-parametrization (https://en.wikipedia.org/wiki/Feynman_parametrization), which leads to integrals from $0$ to $1$ with integrands similar to $I$. Then one inserts the Mellin-Barnes representation and can perform the outer integrations trivially. Only the Mellin-Barnes integrals are left to perform. In my case it is a bit more complicated and I cannot use Feynman parametrization.
EDIT Addition: Stumbling upon the mellin transformation representation of exp(-y) I noticed the same problem here:
$$\int_0^\infty \mathrm{d}y\, e^{-y} = 1 = \frac{1}{2\pi i}\int_0^\infty \mathrm{d}y\int_{c-i\infty}^{c+i\infty}\mathrm{d}s\, \Gamma(s) y^{-s}$$
Best Answer
The problem lies with the Mellin-Barnes representation $$ \frac{1}{(a+b)^\lambda} = \frac{1}{\Gamma(\lambda)}\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \mathrm{d}z\,\Gamma(-z)\Gamma(\lambda+z)\frac{a^z}{b^{\lambda+z}} $$ You can close the contour either to the right or to the left, depending on whether a is larger or smaller than b. However, when you are integrating over $\alpha$, in one part it is smaller than a and in the other part larger, so there's no way to close the contour that would be always valid. What you can do is split the integration over $\alpha$ in two parts, and exchange the roles of a and b in each part, viz. $$ \int_0^\infty \frac{1}{(a+\alpha)^\lambda} \mathrm{d} \alpha = \int_0^a \frac{1}{\Gamma(\lambda)}\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \mathrm{d}z\,\Gamma(-z)\Gamma(\lambda+z)\frac{\alpha^z}{a^{\lambda+z}} \mathrm{d} \alpha \\+ \int_a^\infty \frac{1}{\Gamma(\lambda)}\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \mathrm{d}z\,\Gamma(-z)\Gamma(\lambda+z)\frac{a^z}{\alpha^{\lambda+z}} \mathrm{d} \alpha $$ This is then convergent for $a>0, \lambda > 1$ (taking e.g. $c=\lambda/2$), and you can interchange the $\alpha$ and the $z$ integration.