The book "Table of Integrals, Series, and Products" by Gradshteyn and Ryzhik contains the following two integrals : ( 7th edition section 3.481 ):
- $$\int_{-∞}^∞ xe^x \exp (−μe^x)\,dx =\frac{−1}{μ }(C + \ln μ)$$
- $$\int_{-∞}^∞ xe^x \exp (−μe^{2x})\,dx =\frac{−1}{μ} [C+\ln(4μ)]\sqrt{π/μ} $$
$C$ is Euler's constant
For $[\operatorname{Re} μ > 0]$
I want to generalize these type of integrals to :
$$I(n;t) = \int_t^∞ x^n e^{μ_1x} \exp (−μ_2e^{2x}) \, dx $$
For cases : $ t=0, -\infty$
Question : How to evaluate $I(n;t)$ for a closed form?
For $t=-\infty$: $$ \int_{-\infty}^{ + \infty } {x^n e^{\mu _1 x} \exp ( - \mu _2 e^{2x} )dx} = \frac{1}{{2^{n + 1} }}\int_0^{ + \infty } {(\log t)^n t^{\mu _1 /2 - 1} \exp ( - \mu _2 t)dt} . $$ But since $$ \int_0^{ + \infty } {t^{z/2 - 1} \exp ( - wt)dt} = \frac{{\Gamma (z/2)}}{{w^{z/2} }}, $$ one has $$ \frac{1}{{2^n }}\int_0^{ + \infty } {(\log t)^n t^{z/2 - 1} \exp ( - wt)dt} = \frac{{d^n }}{{dz^n }}\frac{{\Gamma (z/2)}}{{w^{z/2} }}. $$ Hence, $$ \int_{-\infty}^{ + \infty } {x^n e^{\mu _1 x} \exp ( - \mu _2 e^{2x} )dx} = \frac{1}{2}\left[\frac{{d^n }}{{dz^n }}\frac{{\Gamma (z /2)}}{{\mu _2^{z /2} }}\right]_{z=\mu_1}. $$ You may expand the right-hand side via the Leibniz formula.