1. ## Integration problem

Can anyone help me to get $S(x)$ from

$
S^{'} (x)=x^{-b} e^{ -a x}
$

$x>0$

If I tryed to use integration by parts, by I can't figure put, what to call $f$ and $g$. And Maple does not give me anything to go with.

2. Originally Posted by Mourits
Can anyone help me to get $S(x)$ from

$
S^{'} (x)=x^{-b} e^{ -a x}
$

$x>0$

If I tryed to use integration by parts, by I can't figure put, what to call $f$ and $g$. And Maple does not give me anything to go with.
If it's a definite integral (from $0$ to $\infty$), then the answer is $a^{b-1}\Gamma(1-b)$, and I can show that, but I'm not sure how to solve the indefinite integral to find $S(x)$. Wolfram Alpha gives this answer, but I don't know how to get there at the moment.

3. Originally Posted by redsoxfan325
If it's a definite integral (from $0$ to $\infty$), then the answer is $a^{b-1}\Gamma(1-b)$, and I can show that, but I'm not sure how to solve the indefinite integral to find $S(x)$. Wolfram Alpha gives this answer, but I don't know how to get there at the moment.
I think your definite integral answer is the one he is after but... As for the indefinite one.

Express $e^{-ax}$ as a power series and integrate etc...

Should end up with...

$x^{b+1} \sum_{n=0}^{\infty} \frac{(-ax)^n}{(b+n+1)n!}$

But I'm not sure how to complete it from there!

4. Supposing we are actually interested in the definite integral... in case Mourits needs to see how to arrive at that answer.

By letting $x=t$ and $a=s$, we can see this is just a Laplace transform:

$\int_0^{\infty} x^{-b} e^{-ax} \, dx = \int_0^{\infty} t^{-b} e^{-st} \, dt = \mathcal{L} \{ t^{-b} \}$

Using a Laplace transform table, we get

$\mathcal{L} \{ t^{-b} \} = \frac{\Gamma(-b+1)}{s^{-b+1}} = \frac{\Gamma(1-b)}{a^{1-b}} \qquad \mbox{for }b<1 \mbox{ and } a>0$

5. Thanks