<s>[INST] Proving a Difficult Definite Integral in One Variable Let $t > 0 $, $N \in \mathbb{N}$, $q \geq 1 + \frac{1}{N}$.
Also let $r > 0 $, $r' = \frac{r}{r-1}$, such that $\frac{N}{2}(1 - \frac{1}{r}) + \frac{1}{2} < 1$.
We wish to prove the following integral equation:
$ \displaystyle \int^{t}_{0} \large (t-s)^{ -\frac{N}{2}(1 - \frac{1}{r}) - \frac{1}{2} } (t+s)^{ - \frac{N}{2} (q - \frac{1}{r'}) } \text{d}s = \large C t^{  \frac{1}{2} - \frac{N}{2} q } $,
where $C > 0$ is a constant, possibly just $1$.
I'm afraid I do not even know where to begin with this monster. Integral calculators have not been of much help. I presume it is a key point that $\frac{N}{2}(1 - \frac{1}{r}) + \frac{1}{2} < 1$, as the negative of this figure appears as the exponent of $(t-s)$.
Any hints/suggestions as to how I can calculate this integral are much appreciated.
EDIT
After substituting $s = tx$, we arrive at the integral:
$ \large t^{\frac{1}{2} - \frac{N}{2}q }  \int^{1}_{0} (1-x)^{-\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} } (1+x)^{ -\frac{N}{2} (q - \frac{1}{r'}) } \text{d}x $.
Thus, it remains only to show that 
$\int^{1}_{0} (1-x)^{-\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} } (1+x)^{ -\frac{N}{2} (q - \frac{1}{r'}) } \text{d}x = C$.
From our setting of $r$, as explained above, we have the following bounds on each of the exponents, which seem to be important for the existence of this integeral:
$ \large -1 < -\frac{N}{2} (1 - \frac{1}{r}) - \frac{1}{2} < 0 $
$ \large -\frac{1}{2} -\frac{N}{2}q < -\frac{N}{2} (q - \frac{1}{r'})  < \frac{1}{2} -\frac{N}{2}q  $
In particular, by our definition of $q$, both exponents are always negative.
 [/INST] To show that the integral $$\int_0^1(1-x)^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}(1+x)^{-\frac{N}{2}(q-\frac{1}{r'})}\,dx$$ converges, note that $1\leq 1+x\leq 2$ in $(0,1)$, therefore the term $(1+x)^{-\frac{N}{2}(q-\frac{1}{r'})}$ is bounded above by a constant $C$. Therefore, \begin{align*}\int_0^1\left|(1-x)^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}(1+x)^{-\frac{N}{2}(q-\frac{1}{r'})}\right|\,dx&\leq C\int_0^1(1-x)^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}\,dx\\ &=C\int_0^1y^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}\,dy,\end{align*} and the last integral converges, since the exponent of $y$ is greater than $-1$.
</s>