Calculating the area under the curve: $\frac{\mu^x\cdot e^{-\mu}}{Γ(x+1)}$

Question

I am trying to calculate the area under the curve for the function:

Formula 1: $$\frac{\mu^xe^{-\mu}}{Γ(x+1)}$$

Where:

1. $x = 0$ and $x = 1$
2. $μ = 0.5$

The Indefinite Integral equation WolframAlpha gives me is:

Formula 2:

$$\int \frac{μ^xe^{-μ}}{Γ(x+1)}\,dx = \frac{e^{-μ} \operatorname{Ei}((x+1)\ln(μ))}{Γ(μ)}$$

(see this link )

When I solve this equation for $x = 1$ and $x = 0$, and subtract the latter from the former, I end up with $0.088975$ area units.

Having an area value so small between $x = 0$, and $x = 1$ makes no sense to me.

Just by looking at the curve in excel, I can see that when $x = 0$, $y = 0.606$. And when $x = 1$, $y = 0.303$. The area should be approximately $75\%$ of a rectangle with dimensions $0.606 \cdot 1 = 0.606$ area units.

Any help/insight would be appreciated!

Answer 1

I am quite skeptical about a possible closed form of the antiderivative $$I=e^{-μ}\int \frac{μ^x}{\Gamma(x+1)}\,dx$$

However, assuming that we should always stay in the range $0 \leq x \leq 1$, we could make some approximations representing $\Gamma(x+1)$ as a more or less constrained polynomial in $x$

$$\Gamma(x+1) \sim \sum_{n=0}^p a_n x^n=a_p \prod_{n=1}^p (x-r_n)$$ Then partial fraction decomposition would lead to $$I=\frac {e^{-μ}}{a_p} \sum_{n=1}^p b_n\int \frac{\mu^x}{x-r_n}\,dx$$ $$J_n=\int \frac{\mu^x}{x-r_n}\,dx=\mu ^{r_n}\, \log (\mu)\,\text{Ei}\big[x-r_n\big] $$

Trying for $\mu=0.5$ and a few values of $p$ for the integral between $0$ and $1$, the following results were obtained (no constraint for the curve fit) $$\left( \begin{array}{cc} p & \text{approximation} \\ 2 & 0.47485397 \\ 3 & 0.47487261 \\ 4 & 0.47487351 \\ 5 & 0.47487387 \\ 6 & 0.47487409 \end{array} \right)$$ while the exact value is $0.47487382$.

Edit

Another possibility is to use $$\frac{1}{\Gamma(x+1)}=\sum_{n=0}^p \frac{c_n}{n!}\,x^n +O(x^{p+1})$$ where the very first coefficients are (have a look here) $$\left( \begin{array}{cc} 0 & 1 \\ 1 & \gamma \\ 2 & \gamma ^2-\frac{\pi ^2}{6} \\ 3 & \gamma ^3-\frac{\gamma \pi ^2}{2}+2 \zeta (3) \\ 4 & \gamma ^4-\gamma ^2 \pi ^2+\frac{\pi ^4}{60}+8 \gamma \zeta (3) \\ 5 & \gamma ^5-\frac{5 \gamma ^3 \pi ^2}{3}+\frac{\gamma \pi ^4}{12}-\frac{10}{3} \left(-6 \gamma ^2+\pi ^2\right) \zeta (3)+24 \zeta (5) \\ 6 & \gamma ^6-\frac{5 \gamma ^4 \pi ^2}{2}+\frac{\gamma ^2 \pi ^4}{4}-\frac{5 \pi ^6}{168}-20 \gamma \left(-2 \gamma ^2+\pi ^2\right) \zeta (3)+40 \zeta (3)^2+144 \gamma \zeta (5) \end{array} \right)$$

$$K_n=\int \mu^x \,x^n \,dx=-x^{n+1} E_{-n}\big[-x \log (\mu )\big]$$

Using the terms given in the link, the results is $0.47487437$.

Update

Another possibility is based on the fact that$$\frac{1}{\Gamma(x+1)}-1$$ looks like a Gibbs enxcess energy model.

So, let use write $$\frac{1}{\Gamma(x+1)}=1+x(x-1) \sum_{k=0}^p d_k\, x^k$$ This would make

$$I=e^{-μ}\Big[\frac{\mu ^x}{L} +\sum_{k=0}^p (-1)^k d_k L^{-k}\,\left(\frac{\Gamma (k+3,-x L)}{L^3}+\frac{\Gamma (k+2,-xL ))}{L^2} \right)\Big]$$ where $L=\log(\mu)$.

For $p=3$, we should have $$d_0=-\gamma\qquad d_1=-\gamma -\frac{\gamma ^2}{2}+\frac{\pi ^2}{12}$$ $$d_2==48+16 \gamma +2 \gamma ^2-\frac{\pi ^2}{3}+\frac{16 (-8+\gamma +2\log (2))}{\sqrt{\pi }}$$ $$d_3=\frac{384-6 (32+\gamma (10+\gamma )) \sqrt{\pi }+\pi ^{5/2}+96 (2-\gamma -2\log (2))}{3 \sqrt{\pi }}$$ which reproduce the function and first derivative values at $x=0$, $x=\frac 12$ and $x=1$.

For the test example, the result would be $0.47487809$