# Show that Γ(x+1) = exp(-x)*x^(1/2)*

## Homework Statement

I need to prove the relation:

$$\Gamma(x+1) = e^{-x}x^{x+1/2}\int_{-\sqrt{x}}^\infty e^{-\sqrt{x}t}\left (1 +\frac{t}{\sqrt{x}}\right )^x\,dt \qquad(1)$$

## Homework Equations

Definition of Gamma function ?

$$\Gamma(x) = \int_0^\infty t^{x - 1}e^{-t}\,dt$$

## The Attempt at a Solution

I have NO idea how to start this. I figured I should start out with the definition:

$$\Gamma(x+1)=\int_0^\infty t^{x}e^{-t}\,dt \qquad(2)$$

But this does not do much. I tried letting u2 = t -->dt = 2u*du and putting that back into (2):

$$\Gamma(x+1) = 2\int_0^\infty u^{2x}e^{-u^2}u\,du \qquad(3)$$

$$t = 1 + \frac{u}{\sqrt{x}}.$$

$$t = 1 + \frac{u}{\sqrt{x}}.$$

Hi fzero I will try this, but may I ask how you decided on this substitution? I am trying to learn how to approach these, but it is proving to be quite difficult. Thanks.

The integration variables in both formulae are linearly related (by comparing the exponents of the exponentials), so comparing the expressions that appear raised to the $$x$$ power give the relation up to possible constant factors.

I seem to be making progress, but I am again getting stumped; using your suggestion:

$$t = 1 + \frac{u}{\sqrt{x}}\Rightarrow dt = \frac{du}{\sqrt{x}}\qquad(4)$$

plugging into (2):

$$\Gamma(x+1) = \int_{u = -\sqrt{x}}^\infty (1+\frac{u}{\sqrt{x}})^x* e^{-(1+\frac{u}{\sqrt{x}})}\,\frac{du}{\sqrt{x}}\qquad(5)$$

If I let $\alpha = (1+u/\sqrt{x})\Rightarrow d\alpha = du/\sqrt{x}$ and hence

$$\Gamma(x+1) = \int_0^\infty \alpha^xe^{-\alpha}\,d\alpha \qquad(6)$$

:rofl: Wow ... I think I am back to the start now! I have worked in a complete circle ... Doh!

Oh, the "constant" factor I mentioned got me, sorry. Try

$$t = x + \sqrt{x} u$$

Oh, the "constant" factor I mentioned got me, sorry. Try

$$t = x + \sqrt{x} u$$

Ha! Ok .. I'll try this one fzero, I am sorry, I don't really understand your explanation of how you are choosing these substitutions. We are trying to show that

$$\Gamma(x+1)=\int_{t=0}^\infty t^{x}e^{-t}\,dt = e^{-x}x^{x+1/2}\int_{t = -\sqrt{x}}^\infty e^{-\sqrt{x}t}\left (1 +\frac{t}{\sqrt{x}}\right )^x\,dt \qquad(7)$$

So what you are saying is that there is some sort of relationship between the bounds of the integral on the left and the integral on the right ... correct? Should I change the dummy variable in one of the integrals to make a distinction? Or are they both actually 't' ?

Also, if I make your substitution $t = x + \sqrt{x} u$ then $t\rightarrow\infty \quad u\rightarrow\infty \text{ and }t\rightarrow0 \quad u\rightarrow-x/\sqrt{x}$ which is not the lower bound on the integral on the right.

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This is ridiculous. I feel like your original substitution should work. Let's back up a little here. It is being claimed that

$$\int_{t=0}^\infty t^{x}e^{-t}\,dt = e^{-x}x^{x+1/2}\int_{u = -\sqrt{x}}^\infty e^{-\sqrt{x}u}\left (1 +\frac{u}{\sqrt{x}}\right )^x\,du \qquad(8)$$

Now I believe that what you were trying to say is that if we compare coefficients of like powers of x, we might infer a substitution. So we might say that $t = 1+u/\sqrt{x}\Rightarrow dt = du\sqrt{x}$ would make a good substitution. Doing so yields:

$$\int_{u = -\sqrt{x}}^\infty (1+\frac{u}{\sqrt{x}})^x* e^{-(1+\frac{u}{\sqrt{x}})}\,\frac{du}{\sqrt{x}}\qquad (9)$$

Which is getting much closer to the right-hand-side of (8). As a matter of fact, if we equate the RHS of (8) to (9), we have:

$$e^{-x}x^{x+1/2}\int_{u = -\sqrt{x}}^\infty e^{-\sqrt{x}u}\left (1 +\frac{u}{\sqrt{x}}\right )^x\,du = \int_{u = -\sqrt{x}}^\infty (1+\frac{u}{\sqrt{x}})^x* e^{-(1+\frac{u}{\sqrt{x}})}\,\frac{du}{\sqrt{x}} \qquad(10)$$

So I need to show that this is an identity. I start by integrating the RHS of (10) by parts. Letting $u^* = (1 + u/\sqrt{x})^x\Rightarrow du^* = x(1 + u/\sqrt{x})^{x-1}/\sqrt{x}\,du$ and let $dv = e^{-(1+u/\sqrt{x})}du\Rightarrow v = -\sqrt{x}e^{-(1+u/\sqrt{x})}$

Then our integral becomes:

$$I = u^*v - \int v\,du^* = \left [\frac{1}{\sqrt{x}}*(1 + \frac{u}{\sqrt{x}})^x*-\sqrt{x}e^{-(1+\frac{u}{\sqrt{x}})}\right ]_{u=\sqrt{x}}^\infty \quad - \quad \frac{1}{\sqrt{x}}\int_{u=\sqrt{x}}^\infty (1+\frac{u}{\sqrt{x}}^{x-1}\,du\qquad(11)$$

$$\Rightarrow I = x\int_{u=\sqrt{x}}^\infty \sqrt{x}e^{-(1+\frac{u}{\sqrt{x}})}(1 + \frac{u}{\sqrt{x}})^{x-1}\,du \qquad(12)$$

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Also, if I make your substitution $t = x + \sqrt{x} u$ then $t\rightarrow\infty \quad u\rightarrow\infty \text{ and }t\rightarrow0 \quad u\rightarrow-x/\sqrt{x}$ which is not the lower bound on the integral on the right.

$$u\rightarrow-x/\sqrt{x} = - \sqrt{x}$$

is correct.

To see where this came from, compare the exponentials,

$$e^{-t} \rightarrow e^{-x - \sqrt{x} u}.$$

$$u\rightarrow-x/\sqrt{x} = - \sqrt{x}$$

is correct.

To see where this came from, compare the exponentials,

$$e^{-t} \rightarrow e^{-x - \sqrt{x} u}.$$

Oh jeesh ... man .. you think (12) is useless then? Probably eh? ...

So why did you compare the exponentials and not, say, the t^x with (1+u/sqrt(x))^x ?

OK. So I have used the substitution $t = x + \sqrt{x} u$ in (2). I have gotten my integral to look like the following:

$$I = e^{-x}\sqrt{x}\int_{-\sqrt{x}}^\infty\,\,\frac{e^{-\sqrt{x}u}}{\sqrt{x}}\frac{x}{\sqrt{x}}\left (\sqrt{x}u+x \right )^{x-1}\,du \qquad(13)$$

I am not sure how to make (13) look like

$$e^{-x}x^{x+1/2}\int_{u = -\sqrt{x}}^\infty e^{-\sqrt{x}u}\left (1 +\frac{u}{\sqrt{x}}\right )^x\,du \qquad(14)$$

any thoughts? I can't even fathom how I might get an xx factor in there ...

Well

$$\left (\sqrt{x}u+x \right )^{x-1} = x^{x-1} \left (1+\frac{u}{\sqrt{x}}\right )^{x-1},$$

but I think you started with the formula for $$\Gamma(x)$$ instead of $$\Gamma(x+1)$$.

So why did you compare the exponentials and not, say, the t^x with (1+u/sqrt(x))^x ?

Well I did the first time, I was just too fast and got the coefficient wrong.

Hello again fzero! Hmmm ... I started with

$$\Gamma(x+1) = \int_0^\infty t^{(x+1)-1}e^{-t}\,dt = \int_0^\infty t^{x}e^{-t}\,dt \qquad(14)$$

so I think that I started with the right formula.

And how are you finding that

$$\left (\sqrt{x}u+x \right )^{x-1} = x^{x-1} \left (1+\frac{u}{\sqrt{x}}\right )^{x-1},$$

? Can you give me a hint? I am just not seeing it

Hello again fzero! Hmmm ... I started with

$$\Gamma(x+1) = \int_0^\infty t^{(x+1)-1}e^{-t}\,dt = \int_0^\infty t^{x}e^{-t}\,dt \qquad(14)$$

so I think that I started with the right formula.

If that's true, then $$\left (\sqrt{x}u+x \right )^{x}$$ should appear instead of $$\left (\sqrt{x}u+x \right )^{x-1}$$.

And how are you finding that

$$\left (\sqrt{x}u+x \right )^{x-1} = x^{x-1} \left (1+\frac{u}{\sqrt{x}}\right )^{x-1},$$

? Can you give me a hint? I am just not seeing it

Just factor out $$x$$ from the expression inside the parentheses. I probably made it look a bit confusing by swapping terms around the + sign.

If that's true, then $$\left (\sqrt{x}u+x \right )^{x}$$ should appear instead of $$\left (\sqrt{x}u+x \right )^{x-1}$$.

Just factor out $$x$$ from the expression inside the parentheses. I probably made it look a bit confusing by swapping terms around the + sign.

Hi fzero! The (x - 1) occurs when I integrate by parts along the way... Perhaps I do not need to actually integrate?

Hi fzero! The (x - 1) occurs when I integrate by parts along the way... Perhaps I do not need to actually integrate?

You don't need to integrate by parts. All that does is show that $$\Gamma(x+1) = x \Gamma(x)$$, which is important but not directly relevant to this identity.

Thanks for your patience fzero! I have solved it; I made it more difficult by trying to integrate when I did not have to. I posted the solution below. I would like to go back to something for a moment if you have time to talk about it. You guessed the substitution by comparing the exponents on 'e' but you also made it seem like you could have guesses them by comparing the 'bases' being raised to the 'x.'

So in the same way that I needed to put the factor of e-x back inside of the integral to get the whole story --> e-(√(x)u + x) and then comparing to e-t I also could have put the factor of
xx back inside of the integral to get (√(x)u + x)x which could be compared to tx to make the same inference.

I guess I do understand how you got the substitution now, which is much more important to me then the answer to this problem.