# An Integral Problem

Discussion in 'Physics & Math' started by Facial, Mar 16, 2016.

1. ### FacialValued Senior Member

Messages:
2,220
In the middle of an engineering problem (fatigue modeling), I sought Gradshteyn and Ryzhik to solve the following:

$\int \limits_0^a x^\alpha e^{\beta x - x^2}~dx$

It looks deceptively simple and two nights of work on this didn't yield any useful results. Is it only solvable by Mathematica? One of my coworkers found a solution with hypergeometric series but, is there a simpler solution?

Alpha = real
Beta = real
a = positive real

3. ### rpennerFully WiredValued Senior Member

Messages:
4,833
I don't see why you think that this would be simple. You have a product, so the first thing that comes to mind is substitution.

So lets use $x = u + \frac{\beta}{2}$ to get $u = x - \frac{\beta}{2} , \; du = dx, \; -u^2 = -\frac{\beta^2}{4} + \beta x - x^2$.

So we have
$\int \limits_0^a x^\alpha e^{\beta x - x^2} \, dx = e^{\frac{\beta^2}{4}} \int \limits_{- \frac{\beta}{2}}^{a- \frac{\beta}{2}} \left( u + \frac{\beta}{2} \right)^\alpha e^{- u^2} \, du$

Now by using the generalized binomial theorem and assuming $\alpha$ is not a negative integer and $0 < 2 a < \left| \beta \right|$ we have:
$\left( u + \frac{\beta}{2} \right)^\alpha = \sum \limits_{k=0}^{\infty} \frac{ \beta^{\alpha -k} u^k }{k! 2^{\alpha -k} } \prod \limits_{j=0}^{k-1} (\alpha -j) = \sum \limits_{k=0}^{\infty} { \alpha \choose k } \frac{ \beta^{\alpha -k} u^k }{2^{\alpha -k}}$

So
$\int \limits_0^a x^\alpha e^{\beta x - x^2} \, dx \\ = e^{\frac{\beta^2}{4}} \int \limits_{- \frac{\beta}{2}}^{a- \frac{\beta}{2}} \left( u + \frac{\beta}{2} \right)^\alpha e^{- u^2} \, du \\ = e^{\frac{\beta^2}{4}} \sum \limits_{k=0}^{\infty} { \alpha \choose k } \frac{ \beta^{\alpha -k} }{2^{\alpha -k}} \int \limits_{- \frac{\beta}{2}}^{a- \frac{\beta}{2}} u^k e^{- u^2} \, du$

And these integrals we can do with the incomplete Gamma function:

$\int u^k e^{- u^2} \, du = - \frac{1}{2} \Gamma \left( \frac{1 + k }{2} , \, u^2 \right) + C$

Looks nice, but it's not. We get some extra minus signs in there for the $u^2$ term and it gets worse if a is too large.

Let's test for $a = \frac{1}{10}, \alpha = 3, \beta = 1$
$\int \limits_0^{\frac{1}{10}} x^3 e^{x - x^2} \, dx \\ = e^{\frac{1}{4}} \sum \limits_{k=0}^{3} { 3 \choose k } 2^{k-3} (-1)^k \int \limits_{- \frac{5}{10}}^{- \frac{4}{10}} u^k e^{- u^2} \, du \\ = \frac{e^{\frac{1}{4}} }{16} \left( \left( \Gamma \left( \frac{1}{2} , \, \frac{16}{100} \right) - \Gamma \left( \frac{1}{2} , \, \frac{25}{100} \right) \right) - 6 \left( \Gamma \left( 1 , \, \frac{16}{100} \right) - \Gamma \left( 1 , \, \frac{25}{100} \right) \right) + 12 \left( \Gamma \left( \frac{3}{2} , \, \frac{16}{100} \right) - \Gamma \left( \frac{3}{2} , \, \frac{25}{100} \right) \right) - 8 \left( \Gamma \left( 2 , \, \frac{16}{100} \right) - \Gamma \left( 2 , \, \frac{25}{100} \right) \right) \right)$

0.000026904852296752657572155912554654375603197730055...

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