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Playing around with the Standard Normal distribution, $\exp\left(-x^2\right)$, I was wondering about generalizing the distribution by parameterizing the $2$ to a variable $a$. After graphing the distribution for different values of $a$, I decided to see how the value of $a$ affected the area under the curve, which I call $f(a) = \int _{-\infty}^{\infty} \exp\left(-|x|^a\right)dx$. Unable to calculate this analytically (the usual method for calculating the case of $a=2$ by polar coordinates does not generalize) I decided to calculate the function numerically. Here is a graph of the function for $a\in[1,10]$:

graph

As you can see, the function has a minimum near (but not at) $a=2$. I am particularly interested in finding the exact value of the minimum value, though any other facts about this function would be interesting to know. Using Python, my best estimate for the global minimum is $a\approx2.1662269$.

Here is a full list of what I know about the function:

  • As $a\to0$, $f(a)\to\infty$
  • As $a\to\infty$, $f(a)\to2$
  • There is a global minimum around $a\approx2.1662269$
  • $f$ is monotonically decreasing on the interval $(0,\mathrm{minimum})$ and monotonically increasing on $(\mathrm{minimum},\infty)$

Besides that I'm stumped. I tried calculating $\frac{d}{da}f(a) = \frac{d}{da}\int _{-\infty}^{\infty} \exp\left(-|x|^a\right)dx$ but I'm not sure how to evaluate that (maybe there's some multivariable generalization of the Fundamental Theorem of Calculus I've forgotten?). I've taken Calculus classes through Calc III and Diff Eq but don't have a clue how to solve this.

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3 Answers 3

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$$\int_0^{\infty} dx \, e^{-x^a} = \frac1{a} \int_0^{\infty} du \, u^{1/a - 1} e^{-u} = \frac1{a} \Gamma \left ( \frac1{a} \right ) = \Gamma \left ( 1+\frac1{a} \right )$$

To find a local minima, use $\Gamma'(x) = \Gamma(x) \psi(x) $. Other than that, you are more or less stuck with a transcendental equation. Certainly, $\psi(x)$ has a zero in the vicinity of the minimum you have pointed out, but I do not have a good, non-numerical way of locating it.

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Continuing from Ron Gordon's answer, we just have to locate the absolute minimum of the function $\Gamma(1+z)$ over $\mathbb{R}^+$. Since $\Gamma(1)=\Gamma(2)=1$, it is trivially in the interval $I=(0,1)$. Since: $$\Gamma'(1+z) = \Gamma(1+z)\,\psi(z+1)\tag{1}$$ we just have to find the only solution in $I$ of: $$ \gamma = \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right).\tag{2}$$ The RHS is a smooth, increasing, concave function over $I$, so just a few steps of Newton's method with starting point $z=\frac{1}{2}$ give that the solution is about $0.461632$, so the minimum of $\Gamma(z+1)$ over $\mathbb{R}^+$ is about $\color{red}{0.8856}$.

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Continuing from Ron Gordon's answer, $$\frac {d}{da}\Gamma(1+\frac 1a)=-\frac{\Gamma \left(1+\frac{1}{a}\right) \psi \left(1+\frac{1}{a}\right)}{a^2}$$ and so, we look for the zero of $\psi \left(1+\frac{1}{a}\right)$.

Taking into account your result, we can expand $\psi \left(1+\frac{1}{a}\right)$ as a Taylor series built at $a=2$ and obtain $$\psi \left(1+\frac{1}{a}\right)=\psi \left(\frac{3}{2}\right)+\left(1-\frac{\pi ^2}{8}\right) (a-2)+\frac{1}{32} (a-2)^2 \left(-16+2 \pi ^2+\psi ^{(2)}\left(\frac{3}{2}\right)\right)+\frac{1}{384} (a-2)^3 \left(192-12 \pi ^2-\pi ^4-12 \psi ^{(2)}\left(\frac{3}{2}\right)\right)+O\left((a-2)^4\right)$$ Numerically $$\psi \left(\frac{3}{2}\right)\approx 0.036489973978576520559$$ $$\psi ^{(2)}\left(\frac{3}{2}\right)\approx -0.82879664423431999560$$ Limiting to the second order, solving the quadratic equation leads to $$a \approx 2.16699$$ Using the third order would lead to $$a \approx 2.16618$$ quite close to your observation which is the perfect result.

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