# Substitution changes limits of integration?

1. Jan 12, 2012

### TUMath

I'm working through rewriting the gamma function as an infinite product, but my question is just about a specific substitution that was made in my textbook. They took the equation:

$$\Gamma_ n(z)=\int_0^n t^{z-1} (1-\frac{t}{n}) ^ndt$$ for Re(z)>0 and n greater than or equal to 1.

and made the substitution s=t/n. The given result is:

$$\Gamma _n(z)=n^z\int_0^1 s^{z-1}(1-s)^nds$$

My question is, why did the upper limit of integration change from n to 1?

Last edited: Jan 12, 2012
2. Jan 12, 2012

### JG89

Put t = n into your equation s = t/n and you get 1.

3. Jan 12, 2012

### TUMath

I don't see how that would change the limit of integration. In order to make the substitution, I did the following:

$$t^{z-1} = n^{z-1} (\frac{t}{t})^{z-1} = n^{z-1} s^{z-1}$$

$$(1-\frac{t}{n})^n = (1-s)^n$$

$$ds=\frac{1}{n}dt \rightarrow dt=nds$$

All that gives me $\Gamma _n(z)=n^z\int_0^n s^{z-1}(1-s)^nds$

I just can't figure out why I would need to change the limit of integration from n to 1.

4. Jan 12, 2012

### JG89

I don't mean to sound flippant with this, but have you not used the substitution rule before in integrals?

For example, if I wanted to integrate $f(x) = 2x(x^2 + 1)^10$ from 1 to 2, then I would use the substitution $u(x) = x^2 + 1$, right? To find the new limits of integration, I would plug evaluate u(1) and u(2), and these would be my new limits of integration.

So s(t) = t/n is your substitution, right? You were integrating from 0 to n, so s(0) = 0 and s(n) = 1. So these are your new limits of integration.

If you aren't familiar with this, I suggest reading up a proof of the substitution rule to see why this is true.

5. Jan 12, 2012

### TUMath

Thanks JG89. It's just been a while. That makes sense.