# Math Help - Finding improper integral

1. ## Finding improper integral

Integral

I=\int t^{2/3} \exp^{a t} dt, where a < 0

2. Originally Posted by Konstantin
I=\int t^{2/3} \exp^{a t} dt, where a < 0

Perhaps you meant $I=\int_{0}^{+\infty}t^{2/3}e^{at}\;dt$ . In such case, use the substitution $u=-at$ and the Gamma function.

3. Originally Posted by Konstantin
Integral

I=\int t^{2/3} \exp^{a t} dt, where a < 0
If the integral is from $0$ to $+ \infty$, then setting $s=-a$ You have...

$\displaystyle \int_{0}^{+\infty} t^{\frac{2}{3}}\ e^{-s t}\ dt = \mathcal{L} \{t^\frac{2}{3}\}= \frac{\Gamma(\frac{5}{3})}{s^{\frac{5}{3}}}$ (1)

Kind regards

$\chi$ $\sigma$

4. The problem that its limits not from 0 to +\infty. Actually, it is a solution of some eqation, depending on coordinate z, in the form

I(z)=\int_0^z t^{2/3} \exp^{at} dt

5. Originally Posted by Konstantin
The problem that its limits not from 0 to +\infty. Actually, it is a solution of some eqation, depending on coordinate z, in the form

I(z)=\int_0^z t^{2/3} \exp^{at} dt
In that case the integral is not an elementary function but it can be represented as Taylor series...

$\displaystyle I(z)= \int_{0}^{z} t^{\frac{2}{3}} e^{a t} dt = \sum_{n=0}^{\infty} \frac{a^{n}}{n!}\ \int_{0}^{z} t^{n+\frac{2}{3}}\ dt= z^{\frac{5}{3}}\ \sum_{n=0}^{\infty} \frac{a^{n}}{(n+\frac{5}{3})\ n!}\ z^{n}$

Kind regards

$\chi$ $\sigma$

6. And it is NOT, for any finite z, an "improper" integral!