Integral $\frac{2}{876} \int_{0}^{T} \left[ x^{-3/8} (1+b\,x)^{3/4} (1+ab\,x)^{-3} e^{(c\,x^2)}+\log{(5/4)}\right]\,\mathrm dx$ I am trying to evaluate definite integral:
$$\frac{2}{876} \int_{0}^{T} \left[ x^{-3/8} (1+b\,x)^{3/4} (1+ab\,x)^{-3} e^{(c\,x^2)}+\log{(5/4)}\right]\,\mathrm dx$$
$a$, $b$, $c$ are all positive ($a=3.8/3777$, $b=3777$, $c=4.8\times 10^{-3}$). I have plotted the whole integrand. In the plot $\tau$ corresponds to $x$ in the integral. The parameter $\eta$ can be ignored. I have tried the Leibniz method (also called Feynman's integration trick) but without any success. Maybe with a clever change of variable it can be expressed in terms of (combinations of) special functions... or at least get some bounds? Actually I have a different way to estimate the integral, since it's a physics problem and I can guess the (upper limit) of it. T should be $T\sim 57$ (months), I hope. Thanks.


Thanks a lot for the insight. The integral does have an analytical solution, but it is not very helpful, because I need to evaluate "chunks" of it. The solution (for completeness) is
$$\begin{align}
f(T)    & = \sum_{n=0}^\infty\frac{c^n T^{2n+5/8}}{n!}\frac{\Gamma(2n+5/8)}{\Gamma(2n+13/8)}F_1\left(2n+\frac{5}{8},-\frac{3}{4},3,2n+\frac{13}{8};-bT,-aT\right)\nonumber\\
        & = \sum_{n=0}^\infty \frac{c^n}{2n+5/8}\frac{T^{2n+5/8}}{n!}F_1\left(2n+\frac{5}{8},-\frac{3}{4},3,2n+\frac{13}{8};-bT,-aT\right)
\end{align}
$$
This was obtained by expanding the exponential function as a power series, changing the variables in order for the limits of the integral to be 0 and 1 and using the integral representation of the first Appell hypergeometric series, with the note that while these expression hold only for specific arguments, they can be analytically extended to larger domains.
However, I agree that one needs to evaluate it numerically to obtain specific results at given times. Thanks!
