How to solve this integral? (something to do with a beta distribution?)

[Ad VanderVen]

TL;DR Summary

Solving the integral (in Maple notation): Int(exp(c[0]*ln(y)/a[0]+c[1]*M*ln(M-y)/a[1]), y = 0 .. M);

I have the following integral (in Maple notation):

Int(exp(c[0]*ln(y)/a[0]+c[1]*M*ln(M-y)/a[1]), y = 0 .. M);

with (in Maple notation):

0<a[0], 0<a[1], 0<c[0], 0<c[1], 0<y, y<M, 0<M.

What is the solution of this integral? I suspect that the solution has something to do with a beta distribution.

 

[pasmith]

Simplifying the integrand and substituting $y = Mt$ yields $$\int_0^M \exp\left( c_0 \frac{\ln(y)}{a_0} + c_1 M \frac{\ln(M - y)}{a_1} \right)\,dy$$
$$= \int_0^M y^{c_0/a_0}(M - y)^{Mc_1/a_1}\,dy =
M^{(c_0/a_0) + (Mc_1/a_1) + 1} \int_0^1 t^{c_0/a_0} (1 - t)^{c_1/a_1}\,dt
$$ which is indeed a beta function.

 

[Ad VanderVen]

Thank you very much. You were of great help.

 

[Ad VanderVen]

If I enter the values $a_{1}\, = \, 0.7$, $a_{0}\, = \, 0.4$, $c_{1}\, = \, 0.9$, $c_{0}\, = \, 0.5$ and $M\, = \,7$ in
$$\int_{0}^{M}\!{y}^{{\frac {c_{0}}{a_{0}}}} \left( M-y \right) ^{{\frac {Mc_{1}}{a_{1}}}}\,{\rm d}y$$
I get $17939559.61$. If I enter the same values in
$${M}^{{\frac {c_{0}}{a_{0}}}+{\frac {Mc_{1}}{a_{1}}}+1}\int_{0}^{1}\! \left( 1-x \right) ^{{\frac {c_{1}}{a_{1}}}}{x}^{{\frac {c_{0}}{a_{0}}}}\,{\rm d}x$$
I get $344821202.1$.

 

[pasmith]

That should be $$\int_0^M y^{c_0/a_0}(M - y)^{Mc_1/a_1}\,dy = M^{(c_0/a_0) + (Mc_1/a_1) + 1} \int_0^1 t^{c_0/a_0} (1 - t)^{Mc_1/a_1}\,dt$$

 

[Ad VanderVen]

Again thank you very much. Sorry for my mistake.