Solve $ \int_0^\infty \exp(-(\rho t)^k)dt$

Question

Let $T$ a random variable that represents a failure time with $S(t)=\exp(-(\rho t)^k)$ where $\rho,k\in\mathbb{R}$. Find the mean survival time of $T$.

I will not give much statistical details about the problem, but what I need to do is find

$$t_m=\int_0^\infty S(t)dt=\int_0^\infty \exp(-(\rho t)^k)dt$$

How I can calculate this integral at hand?

First I think in just take the integrate $\int_{0}^\infty e^{u}du$ where $u=-(\rho t)^k$ but $du=-k\rho(\rho t)^{k-1}$

I checked this integrate in wolfram, but it involves an $erf(x)$ function error function.

and

$$erf(x)=2\Phi(x\sqrt{2})-1$$

but I can't calculate it at hand. Is there a way to solve it without this error function?

Answer 1

That is actually the definition of the Genralised Error Function. Indeed you have:

$$\int_0^{x} e^{-t^n}\ \text{d}t = \sum_{k = 0}^{+\infty} (-1)^k\frac{x^{nk+1}}{(nk + 1)k!}$$

Good reference to start

https://en.wikipedia.org/wiki/Error_function#Generalized_error_functions

In your case, the best way to see the Error Function is

$$\rho t = z ~~~~~~~ \text{d}t = \frac{\text{d}z}{\rho}$$

$$\frac{1}{\rho}\int_0^{+\infty}e^{-z^k}\ \text{d}z$$

With the notable value

$$\text{erf}(\infty) = 1$$

Answer 2

Following @tired suggestion

Let $z=(\rho t)^k$ with $dz=k\rho(\rho t)^{k-1} dt$ then

$$\int_0^\infty \exp(-(\rho t)^k dt=\int_0^\infty e^{-z} \frac{1}{k\rho(\rho t)^{k-1}}dz $$ since $t=z^{1/k}\rho^{-1}$ then

$$\int_0^\infty e^{-z} \frac{1}{k\rho(\rho t)^{k-1}}dz=\frac{1}{k\rho}\int_0^\infty e^{-z}z^{\frac{1}{k}-1}dz=\frac{1}{k\rho}\Gamma(\frac{1}{k})$$