# Integration of factorials

1. Sep 22, 2014

### Nick Jackson

Hello,

well here's my problem: I got this integral and I don't know how to calculate it (I am trying to find if there exists a k that satisfies this relation) :

$$\int_0^k \frac{1}{ ( 4k-4r-2 ) ! ( 4r+1 ) ! }\, \left ( \frac{y}{x} \right )^{4r} dk = \int_0^k \frac{1}{ ( 4k-4r ) ! ( 4r+3 ) ! }\, \left ( \frac{y}{x} \right )^{4r} dk$$

The problem is mainly in the factorial part (they are results of binomial coefficients as you may see)
Any help?

P.S. There probably doesn't exist such a k.

2. Sep 22, 2014

### HallsofIvy

Staff Emeritus
What you have written here is non-sense. For 2 reasons! First is that you cannot have the dummy variable of integration also be a limit of integration. But that is a minor detail- since it is a dummy variable, replace it with some other variable:
$$\frac{\left(\frac{y}{x}\right)^{4r}}{(4r+1)!} \int_0^k\frac{1}{(4p- 4r- 2)!}dp$$$$= \frac{\left(\frac{y}{x}\right)^{4r}}{(4r+ 3)!}\int_0^k\frac{1}{(4p- 4r)!}dp$$

But there is a much more important problem. Whether you call it "k" or "p", the "variable of integration" must be continuous while the factorial is only defined for non-negative integers.

3. Sep 22, 2014

### dextercioby

You could generalize by putting factorial into an Euler Gamma Function, but that would still make no sense.

4. Sep 23, 2014

### Nick Jackson

Sorry I meant to put r in the variable not k. However I see your point about the continuity... I started with having sums in the lhs and the rhs. What do I do now?