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

## Summary:

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

## Main Question or Discussion Point

I have the following integral (in Maple notation):

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

with (in Maple notation):

0<a, 0<a, 0<c, 0<c, 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
Homework Helper
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.

Last edited by a moderator:
Thank you very much. You were of great help.

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

Again thank you very much. Sorry for my mistake.