$$R^*=(\int_{0}^1 \frac{\psi^n}{(1+\lambda\psi^m)}d\psi)^{0.5}$$
Where n, m, and $\lambda$ are all constants.
Assuming that $n-m\ge0$, and $\lambda > 100$, we obtain:
$$R^*=(\int_{0}^1 \frac{\psi^n}{(\lambda\psi^m)}d\psi)^{0.5}=(\frac{1/\lambda}{(n-m+1)})^{0.5}$$
My issue is with this part. Evaluate when $n-m+1=0$ and $n-m+1<0$
For the first part I think we can say that:
$$R^*=(\int_{0}^1 \frac{1}{(\lambda\psi)}d\psi)^{0.5}=ln(1)^{0.5}-ln(0)^{0.5}=\infty$$
And for the second part:
$$R^*=(\int_{0}^1 \frac{\psi^{n-m}}{\lambda}d\psi)^{0.5}=\frac{1^{n-m+1}-0^{n-m+1}}{(n-m+1)\lambda}$$
Which is $\infty$ also.
Apparently there is some way to evaluate this where the result is NOT infinity, but I have no idea what the approach is supposed to be. The only hint I have is that the denominator must approach 1 when concentration ($\psi$) approaches zero, even when $\lambda$ is large in the original expression. Does this mean that I cannot drop 1 from the original expression? In that case the solution is the hypergeometric function, which I'm also pretty sure isn't what I'm supposed to do.
If anyone could point me in the right direction that would be great.
You may not drop anything which would change a "trivial" integral to an improper integral. Would you do it for $\int_0^1 \frac{dx}{1+x}$ ?
The case where $m=n+1$ is quite simple $$\int_{0}^1 \frac{\psi^n}{1+\lambda\,\psi^{n+1}}\,d\psi$$ since the numerator is "almost" the derivative of the denominator. Then, ???
Edit
For the most general case, you can write $$I=\int_{0}^1 \frac{\psi^n}{1+\lambda\,\psi^{n+1}}\,d\psi=\int_{0}^1 {\psi^n}\left(\sum_{k=0}^\infty(-1)^k \lambda ^k \psi ^{k m}\right)\,d\psi=\sum_{k=0}^\infty(-1)^k \frac{ \lambda ^k}{k m+n+1}$$ which can be expressed as $$I=\frac{\, _2F_1\left(1,\frac{n+1}{m};\frac{m+n+1}{m};-\lambda \right)}{n+1}\qquad \text{or} \qquad I=\frac{\Phi \left(-\lambda ,1,\frac{n+1}{m}\right)}{m}$$ where appear the gaussian hypergeometric function and the Hurwitz-Lerch transcendent function.