Evaluating $\int _0^{\infty }e^{x\left(t-p\right)}p^{1-e^{-px}}\:dx$

Question

How do I proceed with this integration? When I try integration by parts, the same thing keeps coming over and over again.

$$I=\int_0^\infty e^{x(t-p)}p^{1-e^{-px}}\,dx$$

Answer 1

Ok lets start evaluating this: $$\int _0^{\infty }e^{x\left(t-p\right)}\:p^{1-e^{-px}}\:dx=p\underbrace{\int _0^{\infty }e^{tx}\:e^{-px}\:p^{-e^{-px}}\:dx}_{u=e^{-px}}$$ $$=\int _0^1u^{-\frac{t}{p}}\:\:p^{-u}\:du=\underbrace{\int _0^1u^{-\frac{t}{p}}\:\:e^{-u\ln \left(p\right)}\:du}_{x=u\ln \left(p\right)}$$ $$\ln \left(p\right)^{\frac{t}{p}-1}\int _0^{\ln \left(p\right)}x^{-\frac{t}{p}}\:\:e^{-x}\:dx$$ As you may notice thats the definition of the lower incomplete gamma function. $$\int _0^xt^{z-1}\:e^{-t}\:dt=\gamma \left(z,x\right)$$ Also lets use the fact that, $$\gamma \:\left(z,x\right)=\Gamma \left(z\right)-\Gamma \left(z,x\right)$$ So your integral, $$\ln \left(p\right)^{\frac{t}{p}-1}\gamma \left(1-\frac{t}{p},\ln \left(p\right)\right)\:=\ln \left(p\right)^{\frac{t}{p}-1}\left(\Gamma \left(1-\frac{t}{p}\right)-\Gamma \left(1-\frac{t}{p},\ln \left(p\right)\right)\right)$$ To finalize lets use this identity involving the exponential integral, $$E_n\left(x\right)=x^{n-1}\Gamma \left(1-n,x\right)$$ So, $$\ln \left(p\right)^{\frac{t}{p}-1}\left(\Gamma \left(1-\frac{t}{p}\right)-\frac{E_{\frac{t}{p}}\left(\ln \left(p\right)\right)}{\ln \left(p\right)^{\frac{t}{p}-1}}\right)$$ Simplifying a bit we arrive at the solution of your integral, $$\boxed{\int _0^{\infty }e^{x\left(t-p\right)}\:p^{1-e^{-px}}\:dx=\ln \left(p\right)^{\frac{t}{p}-1}\Gamma \left(1-\frac{t}{p}\right)-E_{\frac{t}{p}}\left(\ln \left(p\right)\right)}$$