Integral of $\int_0^2(1-z)^{m-1}e^{-\frac{\kappa}{\alpha}z^{\alpha}}dz$

Question

How can I calculate the following integral $?$: $$ \int_{0}^{2}\left(1 - z\right)^{m - 1} \mathrm{e}^{\large -\kappa z^{\alpha}/\alpha} \,\,\mathrm{d}z $$ where $m > 1$, $\kappa > 0$ and $\alpha \in \left(0,2\right]$.

Thanks in advance.

Answer 1

$\int_0^2(1-z)^{m-1}e^{-\frac{\kappa}{\alpha}z^{\alpha}}~dz$

$=\int_0^2\sum\limits_{n=0}^\infty\dfrac{(-1)^n\kappa^nz^{\alpha n}(1-z)^{m-1}}{\alpha^nn!}~dz$

$=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(-1)^n\kappa^nz^{\alpha n}(1-z)^{m-1}}{\alpha^nn!}~dz+\int_1^2\sum\limits_{n=0}^\infty\dfrac{(-1)^n\kappa^nz^{\alpha n}(1-z)^{m-1}}{\alpha^nn!}~dz$

$=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(-1)^n\kappa^nz^{\alpha n}(1-z)^{m-1}}{\alpha^nn!}~dz-\int_0^1\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+n}\kappa^nz^{m-1}(z+1)^{\alpha n}}{\alpha^nn!}~dz$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\kappa^n\Gamma(m)\Gamma(\alpha n+1)}{\alpha^nn!\Gamma(m+\alpha n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+n}\kappa^n{_2F_1}(m,-\alpha n;m+1;-1)}{m\alpha^nn!}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\kappa^n\Gamma(m)\Gamma(\alpha n+1)}{\alpha^nn!\Gamma(m+\alpha n+1)}-\sum\limits_{n=0}^\infty\dfrac{(-1)^{m+n}\kappa^n2^{\alpha n+1}{_2F_1}(1,m+\alpha n+1;m+1;-1)}{m\alpha^nn!}$

Answer 2

This is not an easy standard integral in my opinion. We can try something "numerical" with care.

First of all, noticing that $k > 0$ and the exponential has a minus, we may attack it with a Taylor expansion:

$$\large e^{-k/a\ z^a} \approx \sum_{j = 0}^{+\infty} \frac{1}{j!}\left(-\frac{k}{a} z^a\right)^j$$

Hence arranging the integral we get

$$\sum_{j = 0}^{+\infty} \frac{1}{j!}\left(-\frac{k}{a}\right)^j \int_0^2 (1-z)^{m-1} z^{aj}\ \text{d}z$$

The latter integral is not trivial. Yet if you have a bit od knowledge about Special Functions you may have seen this:

$$\int (1-z)^A z^B\ \text{d}z = \frac{z^{B+1} \, _2F_1(-A,B+1;B+2;z)}{B+1}$$

Which is indeed the HyperGeometric Special Function.

In your case $A = 1-m$ and $B = aj$, also evaluated between $0$ and $2$, that is:

$$\int_0^2 (1-z)^{m-1} z^{aj}\ \text{d}z = \frac{(-1)^{m+1} 2^{a j+m} \, _2F_1\left(1-m,-a j-m;-a j-m+1;\frac{1}{2}\right)}{a j+m}+\frac{(-1)^{m+1} \Gamma (m) \Gamma (-a j-m)}{\Gamma (-a j)}+\frac{\Gamma (m) \Gamma (a j+1)}{\Gamma (a j+m+1)}$$

Where $\Gamma(\cdot)$ denotes the Euler Gamma Function.

At the end of the game you get:

$$\sum_{j = 0}^{+\infty} \left\{\frac{1}{j!}\left(-\frac{k}{a}\right)^j\frac{(-1)^{m+1} 2^{a j+m} \, _2F_1\left(1-m,-a j-m;-a j-m+1;\frac{1}{2}\right)}{a j+m}+\frac{(-1)^{m+1} \Gamma (m) \Gamma (-a j-m)}{\Gamma (-a j)}+\frac{\Gamma (m) \Gamma (a j+1)}{\Gamma (a j+m+1)}\right\}$$

Which is a cool mess.

Handle with care