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

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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.
 
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I would likely begin by stating:

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

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

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

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

And so we now have:

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

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

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

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

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

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

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

Take the nth derivative with respect to $s$

$$\int_0^\infty e^{-sx}x ^n\,dx=\dfrac{n!}{s^{1+n}}$$
 
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? :eek:
 
I like Serena said:
Btw, if we start with the factorial, we can't "just" switch to $\Pi$ can we? :eek:

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