Simplify integral $f(x)=e^{-(\frac{(m+1)}{C_1}+\frac{(n)}{C_2})x}\ln(x+1)dx$

Question

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This question Let $f(x)$ define by $$ f(x)=e^{-(\frac{(m+1)}{C_1}+\frac{(n)}{C_2})x}\ln(x+1) $$ I would like to compute the following integral $$\int_{0}^{\infty}f(x)dx$$

So we use $y=x+1$ and $u=(\frac{(m+1)}{C_1}+\frac{(n)}{C_2})$, the integral becom $$ \int_{1}^{\infty}e^{-u(y-1)}\ln (y)dy=e^u\int_{1}^{\infty}e^{-uy}\ln (y)dy $$ Use the following result $$ \int_{1}^{\infty}e^{-uy}\ln (y)dy=-\frac{1}{u}E_i(-u) $$ SO the final result is $$ -\frac{e^u}{u}E_i(-u) $$ were $E_i$ is the exponential integral. But were i found the integral the result was $$ -\frac{e^u}{u}E_i(u) $$ No "$-$" inside the exponential integral $E_i$? why

This is the Problem

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The refrence [38] definition is

See here [38] defenition

Answer 1

$\mathrm{Ei}$ and $E_i$ are different functions (see https://dlmf.nist.gov/6.2). For $u>0$ your integral is $$ \int_{1}^{\infty}e^{-uy}\ln (y)dy=+\frac{1}{u}E_1(u) $$ Addendum: With int(exp(-u*y)*ln(y),y=1..infinity) with u>0 Wolfram Alpha returns $\frac{\Gamma(0,u)}{u}$ for the integral , which is the same as my result using $\Gamma(0,u) = E_1(u)$ (see http://functions.wolfram.com/GammaBetaErf/Gamma2/27/02/0002/)