Calculate the integral $\int_0^{\infty} \frac{x^{\alpha-1}e^{-x}+1}{x+2}dx$

Question

Calculate the integral $\int_0^{\infty} \frac{x^{\alpha-1}e^{-x}+1}{x+2}dx$

What I have tried:

Split the integral

$$\int_0^{\infty}\frac{x^{\alpha-1}e^{-x}}{x+2}dx+\int_0^{\infty}\frac{1}{x+2}dx$$

$$=\int_0^{\infty}\frac{x^{\alpha-1}e^{-x}}{x+2}dx$$

Taking the substitution $$u = x+2; du =dx; x= u-2$$ and with the change of variables

$$\implies \int_2^{\infty}\frac{(u-2)^{\alpha-1}e^{-u+2}}{u}du$$

Then integrating by parts with the following substitutions $$z = (u-2)^{\alpha-1}; dx = (\alpha-2)(u-2)^{\alpha-2}du; dv = u^{-1}e^{-u+2}; v = e^2Ei(-u)$$

$$\implies \int_2^{\infty}\frac{(u-2)^{\alpha-1}e^{-u+2}}{u}du=\left[(u-2)^{\alpha-1}e^2Ei(-u)\right]_2^{\infty}-(\alpha-1)e^2\int_2^{\infty}(u-2)^{\alpha-2}Ei(-u)du$$ $$\implies \int_2^{\infty}\frac{(u-2)^{\alpha-1}e^{-u+2}}{u}du=-(\alpha-1)e^2\int_2^{\infty}(u-2)^{\alpha-2}Ei(-u)du$$

How do I proceed in simplifying this any further?

Answer 1

$$\int_0^{\infty} \frac{x^{\alpha-1}e^{-x}+1}{x+2}dx>\int_0^\infty\frac1{x+2}=\ln\infty-\ln 2=\infty$$

Answer 2

if you are intrested by the antiderivatives for integer values of $\alpha \geq 1$

$$I_\alpha=\int \frac{x^{\alpha-1}e^{-x}+1}{x+2}dx$$ they write $$I_\alpha= \log (x+2)+(-1)^{\alpha +1}\, 2^{\alpha -1}e^2\, \text{Ei}(-x-2) -e^{-x} P_{\alpha}(x)$$ where the first polynomials are $$\left( \begin{array}{cc} 1 & 0 \\ 2 & 1 \\ 3 & x-1 \\ 4 & x^2+4 \\ 5 & x^3+x^2+6 x-2 \\ 6 & x^4+2 x^3+10 x^2+12 x+28 \\ 7 & x^5+3 x^4+16 x^3+40 x^2+96 x+64 \\ 8 & x^6+4 x^5+24 x^4+88 x^3+280 x^2+528 x+592 \\ 9 & x^7+5 x^6+34 x^5+162 x^4+664 x^3+1960 x^2+3984 x+3856 \\ 10 & x^8+6 x^7+46 x^6+268 x^5+1356 x^4+5392 x^3+16240 x^2+32352 x+32608 \end{array} \right)$$

The trouble comes from the $1$ in numerator (as already mentioned in comments). But $$J_\alpha=\int_0^\infty \frac{x^{\alpha-1}e^{-x}}{x+2}dx=e^2\, \Gamma (\alpha )\, E_{\alpha }(2)$$