How to integrate $\int e^{\alpha t^{\beta}} dt$

Question

So I have this differential equation, i identified it as first-order linear differential equation.

$\frac{dA}{dt}=-(a-bp)-\alpha\beta t^{\beta-1}A$

note: a, b, p, $\alpha, \beta$ are just some parameters

then i found the integrating factor which is $I(t)=e^{\alpha t^{\beta}}$

after multiplied both sides of the differential equation with the integrating factor and integrating both side, i got

$A=-(a-bp)e^{\alpha t^{\beta}}\int e^{\alpha t^{\beta}} dt $

and the next thing is im stuck at integrating $e^{\alpha t^{\beta}}$ because i couldn't find the result. i tried using substitution method but the variable 't' is always there.

a little insight or ideas or hints would help, thank you.

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$For $0\le x\lt\infty$ and assuming $\alpha\lt 0$:

$$\int_0^x e^{\alpha t^\beta}\d t\overset{t=\sqrt[\beta]{-u/\alpha}}=\frac{1}{\beta\cdot|\alpha|^{1/\beta}}\int_0^{|\alpha|\cdot x^\beta}u^{1/\beta-1}\cdot e^{-u}\d u=\frac{1}{\beta\cdot|\alpha|^{1/\beta}}\cdot\gamma(1/\beta,|\alpha|\cdot x^\beta)$$

Where:

$$\gamma(s,x):=\int_0^x t^{s-1}e^{-t}\d t$$

Is the lower incomplete gamma function.

Assuming $\alpha,\beta$ are such that the integral is convergent.

This is the best you can do, and serves at best to wrap it up into a compact notation

For the case $\alpha\gt0$ see Zachary’s answer. There is sadly no nice standard notation (as far as I know) to write the integral in this case.

Answer 2

Essentially, as FShrike showed, you are interested in the integral $$\int_0^R x^{s-1}e^{x}\,dx.$$ While there is no nice closed form for finite $R%$, we can still obtain upper and lower bounds. You didn't include the range of your parameters so I'll assume that $R>1$ and $s>1$. We can use $e^x>x$ for all $x\in\mathbb{R}$ to show that your integral satisfies the following upper bound $$\int_0^R x^{s-1}e^{x}\,dx>\int_0^R e^{(s-1)x} e^x\,dx=\frac{e^{sR}-1}{s}.$$ We now deduce a lower bound of the form $$\int_0^R x^{s-1}e^{x}\,dx > \int_0^1 x^{s-1}\,dx+\int_1^R e^x\,dx \\ = \frac{1}{s}+e^{R}-1. $$ Hence, we have that $$\frac{1}{s}+e^{R}-1 < \int_0^R x^{s-1}e^{x}\,dx < \frac{e^{sR}-1}{s}.$$