How to find $P(X>x)$ when the density is known but the integral does not seem to converge

Question

I am trying to evaluate
$$P(X>x) = \int_x^{\infty } t^{\kappa } \exp{\left(-\rho t^{\alpha\kappa + 1}\right)} \, dt$$ where $\kappa$, $\rho$ and $\alpha$ are all constants. I have tried some substitutions, integrating by parts to solve the integral, but it did not seem to converge.

I also tried to compute the laplace transform, inverse Fourier transform of the density using Mathematica, but it was n't able to work it out.

After having spent 6-7 hours trying to solve this I am still hopelessly stuck.

Any help would be much appreciated.

Answer 1

Did you try the substitution $u=t^{\alpha\kappa+1}$? As far as I see, you get then something like $k\cdot u^{\beta}\exp(-\rho u)$ ($k,\beta$ constants) as integrand, the integral is then similar to the incomplete Gamma function. https://en.wikipedia.org/wiki/Incomplete_gamma_function

Hope that it helps.

Answer 2

As answered by Karl $$\int t^{\kappa } \exp{\left(-\rho t^{\alpha\kappa + 1}\right)} \, dt=-\frac{t^{\kappa +1} \left(\rho t^{\alpha \kappa +1}\right)^{-\frac{\kappa +1}{\alpha \kappa +1}} \Gamma \left(\frac{\kappa +1}{\alpha \kappa +1},t^{\alpha \kappa +1} \rho \right)}{\alpha \kappa +1}$$ where appears the incomplete gamma function.

This can also write $$\int t^{\kappa } \exp{\left(-\rho t^{\alpha\kappa + 1}\right)} \, dt=-\frac{t^{\kappa +1} E_{\frac{(\alpha -1) \kappa }{\alpha \kappa +1}}\left(t^{\alpha \kappa +1} \rho \right)}{\alpha \kappa +1}$$ where appears the exponential integral function.