Integral of factorial function

Question

$$ \mbox{What can we say about the integral}\quad \int_{0}^{a} x!\,{\rm d}x\ ?. $$

Or something like $\displaystyle\int_{0}^{3} x!\ {\rm d}x\ ?$.

Answer 1

This is a very amazing problem. As I said earlier, I did not find any closed form for the integral and then I only performed numerical integrations.

What is surprizing if that $y(a)=\displaystyle\int_{0}^{a} \Gamma(1+x)~ dx$ looks very much to $x(a)=\Gamma(1+a)$ (plot the two curves as a function of $a$).

Taking into account Lucian's comment, I performed a parametric plot and I observed that, for values of $a \gt 1$, $y$ is almost linear with $x$.

A regression $\log(y)=a +b \log(x)$, for the range $2 \leq a \leq 12$, leads to $a=-0.0775761$ and $b=0.949784$ with $R^2=0.999467$.

Answer 2

Using:

Taylor Series of Gamma Function:

$$\int_0^x t! dt= \lim_{c\to\infty} \sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^{m+1}c^{n+1}\int_0^x t^n dt\text E_{-n}(-c(m+1))}{m!n!} $$

to get the following after an index shift:

$$\int_0^x t!dt= \lim_{c\to\infty} \sum_{m,n=1}^\infty\frac{(-1)^mc^nx^n\text E_{1-n}(-cm))}{\Gamma(m)n!}=\lim_{c\to\infty}\sum_{m,n=1}^\infty\frac{(-1)^{m+n}Q(n,-cm)}{\Gamma(m)n} \left(\frac xm\right)^n$$

Shown here with generalized exponential integral $E_v(z)$ and gamma regularized $Q(a,z)$. Hopefully there is a simplification.

Alternatively, we use the exponential integral Ei$(z)$

$$\int x! dx=\int\lim_{t\to\infty}\sum_{n=0}^\infty\frac{(-1)^n t^{x+n+1}}{n!(n+x+1)}dx=\boxed{C+\lim_{t\to\infty}\sum_{n=1}^\infty\frac{(-1)^{n+1}\text{Ei}((x+n)t)}{\Gamma(n)}}$$

Shown here. When truncating the sum, $n$’s upper bound must be much greater than $t$ to converge.

Answer 3

In fact, you want to compute

$$I(a)=\int_{0}^{a} \Gamma(1+x)~ dx=\int_{0}^{a}x\Gamma(x)~ dx$$

Taking into account that

$$\Gamma(x)=\int_{0}^{\infty}y^{x-1}\ e^{-y}~ dy$$

we have

$$I(a)=\int_{0}^{\infty}\ e^{-y}\left (\int_{0}^{a}x y^{x-1}~ dx\right )~ dy$$

The inner integral is easy to calculate

$$\int_{0}^{a}x y^{x-1}~ dx=\frac{y^{a}(a\ln y-1)+1}{y\ln^2 y}$$

The end result

$$I(a)=\int_{0}^{\infty}\frac{y^{a}(a\ln y-1)+1}{y\ln^2 y}e^{-y}~ dy$$

In principle, we are done.

For a particular value of $a$ we can integrate numerically.

But does there exist any analytical expression for $I(a)$ is a separate question.

I'd bet money it doesn't exist.

But I'd be happy if I was wrong.

Answer 4

The factorial function is only defined on the positive integers, so those don't make sense. However, there is a generalization of the factorial called the Gamma function which you might want to check out.