# Simplifying integral of Gauss' hypergeometric function

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EngWiPy
Hello all,

I have this integral, and currently I'm evaluating it using Mathematica numerically, which takes time to be evaluated. Can I write it in a way that the integral has a formula in the Table of Integrals?

$$\int_0^{\infty} F\left(a_1,a_2;a_3;a_4-a_5x\right) e^{-x}\,dx$$

where ##\{a_i\}_{i=1}^5## are constants, and ##F(.,.;.;) = _2F_1(.,.;.;.)## is the Gauss Hypergeometric function.

Thanks

## Answers and Replies

EngWiPy
The last term in my formula is ##a_4-a_5t## not ##-t##! I tried change of variables, but the limits of integral will change, and there is no integral formula for it.

Mentor
It might save some time to use this analytic result and integrate the rest (from the changed limits) numerically, as the function does not change that much in this smaller range.

If the help page only gives a result for the integral from 0 to infinity, then I guess a closed form for arbitrary limits does not exist.

EngWiPy
$$\int_0^{\infty} F\left(a_1,a_2;a_3;a_4-a_5x\right) e^{-x}\,dx$$

Are the coefficients of your hypergeometric series decreasing rapidly ?
##A_n = \frac{ (a_1)_n (a_2)_n} {(a_3)_n n!} ##

How many terms of the series ##\sum_{k=0}^\infty (\int_0^\infty A_k (a_4 - a_5 x)^k e^{-x} dx ) ## would we have to evaluate to get the precision you need ?

EngWiPy
Are the coefficients of your hypergeometric series decreasing rapidly ?
##A_n = \frac{ (a_1)_n (a_2)_n} {(a_3)_n n!} ##

How many terms of the series ##\sum_{k=0}^\infty (\int_0^\infty A_k (a_4 - a_5 x)^k e^{-x} dx ) ## would we have to evaluate to get the precision you need ?

I have ##a_1=2##, ##a_2=1##, and ##a_3=3##, while ##a_4## changes in a loop. ##a_5## is also constant, but set at the beginning of the simulation.