Can the Definite Integral of an Exponential Function be Expressed as a Product?

[Greg]

Show that $\int_0^\infty e^{-nx}x^{s-1}\,dx=\dfrac{\prod(s-1)}{n^s}$

I'm not familiar with the techniques that would be used to solve this.

 

[MarkFL]

I would likely begin by stating:

\(\displaystyle I(n,s-1)=\lim_{t\to\infty}\left(\int_0^t e^{-nx}x^{s-1}\,dx\right)\)

Okay, now, for the integral, I would use IBP, where:

\(\displaystyle u=x^{s-1}\,\therefore\,du=(s-1)x^{s-2}\,dx\)

\(\displaystyle dv=e^{-nx}\,dx\,\therefore\,v=-\frac{1}{n}e^{-nx}\)

And so we now have:

\(\displaystyle I(n,s-1)=\lim_{t\to\infty}\left(-\frac{1}{n}\left[x^{s-1}e^{-nx}\right]_0^t+\frac{s-1}{n}\int_0^t e^{-nx}x^{s-2}\,dx\right)\)

\(\displaystyle I(n,s-1)=\frac{s-1}{n}I(s-2,n)+\frac{1}{n}\lim_{t\to\infty}\left(\frac{x^{t-1}}{e^{nt}}\right)\)

We observe that the limit goes to zero (polynomial over an exponential) and we have:

\(\displaystyle I(n,s-1)=\frac{s-1}{n}I(n,s-2)\)

Now, if we repeat this process another $s-2$ times, we obtain:

\(\displaystyle I(n,s-1)=\frac{(s-1)!}{n^{s-1}}I(n,0)=\frac{(s-1)!}{n^{s-1}}\lim_{t\to\infty}\left(\int_0^t e^{-nx}\,dx\right)=\frac{(s-1)!}{n^s}\)

 

[alyafey22]

Start by

$$\int_0^\infty e^{-sx}\,dx=\dfrac{1}{s}$$

Take the nth derivative with respect to $s$

$$\int_0^\infty e^{-sx}x ^n\,dx=\dfrac{n!}{s^{1+n}}$$

 

[I like Serena]

Let's start with the definition of $\Pi$, which is:
$$\Pi(z)=\int_0^\infty e^{-t} t^z \,dt$$
Now we can substitute $z=s-1$ and $t=nx$.

Btw, if we start with the factorial, we can't "just" switch to $\Pi$ can we?

 

[alyafey22]

Btw, if we start with the factorial, we can't "just" switch to $\Pi$ can we?

No we can't. This is under the assumption we work with natural numbers.