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.
 A: 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 $-1$.
A: Only a partial answer:
For $a>-1,b>-1,x\in[0,1]$, let $$B(x,a+1,b+1):=\int_0^x u^a \cdot (1-u)^b \,\mathrm du$$ denote the Incomplete Beta function. 
For $a,b>0$ and $c>0$, we can give an anti-derivative on $]0,c[$ of the function $$f:[0,c]\to \mathbb R, x\mapsto(c+x)^a\cdot(c-x)^b$$ in terms of the incomplete Beta function: In your case, I would like to choose $a=- \frac{N}{2} (q - \frac{q}{r'})$ and $b=-\frac{N}{2}(1 - \frac{1}{r})$. Also, note that my $c$ is your $t$. I don't see why $a,b>0$ though.
\begin{split}\int f&=\int(c+x)^a\cdot(c-x)^b\,\mathrm dx\\
&\overset{x=2 c y -c}{=}2c\cdot\int (2 c y)^a\cdot(2c\cdot(1-y))^b\,\mathrm dy \\
&=2^{a+b+1}\cdot c^{a+b+1}\cdot B(y,a+1,b+1) +const.\\
&=2^{a+b+1}\cdot c^{a+b+1}\cdot B\left(\frac{c+x}{2c},a+1,b+1\right) +const.
\end{split}
Put more formally, we have for all $x\in]0,c[$, $F'(x)=f(x)$, where $$F: [0,c]\to\mathbb R, x\mapsto 2^{a+b+1}\cdot c^{a+b+1}\cdot B\left(\frac{c+x}{2c},a+1,b+1\right).$$
(Note that, since $c\geq x$, we have $\frac{c+x}{2c}\in[0,1]$.)
By the fundamental Theorem of calculus, we thus have $$\int_0^c f(x)\,\mathrm dx = F(c)-F(0).$$

EDIT: If $a,b$ are not positive you might use that (this was determined by WolframAlpha) $F'(x)=f(x)$ for $x\in]0,c[$, where 
$$F:[0,c]\to\mathbb R, \\x\mapsto \frac{2^b (c - x)^b (c + x)^{1 + a} (1 - x/c)^{-b} {}_2F_1(1 + a, -b, 2 + a, \frac{c + x}{2 c})}{1 + a}.$$
Here, $${}_2F_1(x_1,x_2,x_3,x_4)$$ denotes the Hypergeometric function. 
A word of caution: I know almost nothing about the Hypergeometric function and it seems that you can run into trouble for $x_4=1$ (i.e. when $x=c$ so it is not clear if $F(c)$ is well-defined).
