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

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:

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

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!

• I think you misread $\Gamma(x+1)$ as $\Gamma \cdot (x+1)$ where $\Gamma$ is some constant. Here, $\Gamma()$ is the Gamma function. You have to use Gamma[] in WolframAlpha. Anyway, when $x$ is an integer, the usual way to compute this is integration by parts and with the knowledge that $\Gamma(x) = (x-1)!$. – angryavian Aug 26 '20 at 0:30
• You were right, turns out that there is no Indefinite Integral generated by WolframAlpha for Formula 1. Unfortunately, I started out with the Poisson Distribution formula and intentionally replaced the factorial denominator, x! with the Gamma function so that I could solve the equation for instances where x >= 0 and continuous. What would be your recommendation for calculating the area under the curve for Formula 1 where x >= 0? – the_rubicon Aug 26 '20 at 1:52
• There is really nothing to compute here, since $\int_0^\infty \mu^x e^{-\mu} \, d\mu$ is the definition of $\Gamma(x+1)$. So the integral (with the normalizing factor $1/\Gamma(x+1)$ )is just $1$. (Also, saying $\mu=0.5$ makes no sense if your integral is with respect to $\mu$.) – angryavian Aug 26 '20 at 2:25
• The Wolfram alpha web site integrated with respect to $x$. But I suspect you want to integrate with respect to $\mu$. Then it is called the incomplete Gamma function. – Stephen Montgomery-Smith Aug 26 '20 at 3:46
• I'm sorry, it seems I have a disconnect. I have plotted Formula 1 in excel with increments of of x increasing by 0.1 from 0 to 4. I have a fixed μ = 0.5 value. Perhaps I should label this something else? I'm really just using μ as a fixed numerical constant I adjust to get the desired graph shape (increasing shifts y_max right and simultaneously reduces amplitude). All I want to know is if it's possible to calculate the area under my graph between 0 and any given decimal value of x, say x = 0.7, given a fixed value of 'μ'. And if this is possible, how... Many thanks in advance, J – the_rubicon Aug 26 '20 at 6:07

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 $$0.47487809$$