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I need to prove that the inverse Mellin trasform I am using gives a proper density function, i.e. that it integrates to 1.
Starting point: I need to solve this integral equation
\begin{equation*} \int \pi_{u|\tau_u}(\tau_u^{-1}Q_u|\tau_u)\pi_{\tau_u}(\tau_u) d\tau_u=\int \pi_{s|\tau_s}(\tau_s^{-1}Q_s|\tau_s)\pi_{\tau_s}(\tau_s) d\tau_s \end{equation*}
where the unknown quantity is $\pi_{\tau_s}(\tau_s)$ for $\tau_s>0$. With the symbol $\pi$ I denote probability distributions, i.e. functions that integrate to 1. Both integrals are marginal distributions. \begin{equation*} \begin{split} & Q_u=u'u \sim Gamma\left(\frac{n}{2},\frac{1}{2}\right) \mbox{ a quadratic form}\\ & \tau_u^{-1}Q_u|\tau_u \sim Gamma\left(\frac{n}{2},\frac{\tau_u}{2}\right)\\ & Q_s=s's \sim \sum_{i=1}^{r} Gamma\left(\frac{1}{2},\frac{1}{2\lambda_i}\right) \mbox{ another quadratic form ($\lambda_i$ are eigenvalues)}\\ & \tau_s^{-1}Q_s|\tau_s \sim \sum_{i=1}^{r} Gamma\left(\frac{1}{2},\frac{\tau_s}{2\lambda_i}\right)\\ & \pi_u(Q_u)=\int \pi_{u|\tau_u}(\tau_u^{-1}Q_u|\tau_u)\pi_{\tau_u}(\tau_u) d\tau_u \sim GB2(1, 2b, \frac{n}{2}, a) \mbox{ marginal distribution of $Q_u$}\\ & \pi_s(Q_s)=\int \pi_{s|\tau_s}(\tau_s^{-1}Q^*_s|\tau_s)\pi_{\tau_s}(\tau_s) d\tau_s \mbox{ marginal distribution of $Q_s$} \end{split} \end{equation*}
I use the Mellin transform to solve the integral equation, and I know that \begin{equation*} \widehat{\pi}_{\tau_s}(2-z)=\frac{\widehat{\pi}_{u}(z)}{\widehat{\pi}_{s|\tau_s}(z)} \mbox{ where } \widehat{\pi} \mbox{ is the Mellin transform, and } z=c+id \in \mathbb{C} \end{equation*}
I know both Mellin transforms on the right hand side: \begin{equation*} \begin{split} \widehat{\pi}_u(z)&=(2b)^{z-1}\frac{\Gamma(\frac{n}{2}+z-1)\Gamma(a-z+1)}{\Gamma(\frac{n}{2})\Gamma(a)} \mbox{ Mellin transform of a GB2 density}\\ \widehat{\pi}_{s|\tau_s}(z)&=(2\beta)^{z-1}\sum_{k=0}^\infty a_k \frac{\Gamma(z+\frac{r}{2}+k-1)}{\Gamma(\frac{r}{2}+k)} \end{split} \end{equation*} using the series representation of the density of a quadratic form, where $r$ is the structure matrix rank, $a_0=\prod_{i=1}^r \left(\frac{\beta}{\lambda_i}\right)$, $a_k=(2k)^{-1}\sum_{j=0}^{k-1}b_{k-j}a_j$, $b_k=\sum_{i=1}^r c_i^k$, $c_i=1-\frac{\beta}{\lambda_i} $ and $\beta$ is such that $|c_i|<1$ for all $i$.\
Finally, I use the Mellin inversion formula to get the unknown quantity $\pi_{\tau_s}(\tau_s)$ \begin{equation*} \pi_{\tau_s}(\tau_s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \widehat{\pi}_{\tau_s}(2-z)\tau_s^{-(2-z)} dz=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{\widehat{\pi}_{u}(z)}{\widehat{\pi}_{s|\tau_s}(z)}\tau_s^{-(2-z)} dz \end{equation*}
HOW DO I PROVE THAT THIS DISTRIBUTION IS PROPER, I.E. THAT \begin{equation*} \int_0^\infty \pi_{\tau_s}(\tau_s) d\tau_s=\int_0^\infty\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{\widehat{\pi}_{u}(z)}{\widehat{\pi}_{s|\tau_s}(z)}\tau_s^{-(2-z)} dzd\tau_s=1 \mbox{ ???} \end{equation*}
Please be patient with mistakes/lack of rigour, I am not a mathematician...
Any help will be appreciated! Thank you very much :)