integral involves incomplete gamma function and $x^{-1}$ I am trying to solve this integral
$$\int_z^{\infty } \frac{\Gamma (0,m y)}{y} \, dy$$
 A: Writing out the incomplete Gamma function, you get
$$I=\int_z^\infty \int_{my}^\infty y^{-1}t^{-1}e^{-t}{\rm d}t\,{\rm d}y$$
Drawing the integration area in the $(y,t)$ plane, you get an upper triangle above the line $t=my$ and to the right of $y=z$. Exchanging integration order, you get
$$I= \int_{mz}^\infty \int_z^{t/m} y^{-1}t^{-1}e^{-t}{\rm d}y\,{\rm d}t=$$
$$=\int_{mz}^\infty \ln\frac{t}{mz} t^{-1}e^{-t} {\rm d} t$$
Mathematica gives me a Hypergeometric function, so if that's what you're after, you're done. The nice part here is that $mz$ is the sole parameter of this function (at first it looked like you have two parameters, so this is an improvement). This way, you can also rewrite it with $u=t/mz$ and you get fixed integral bounds, with parameter only remaining in the exponent:
$$I=\int_{1}^\infty \ln u\, u^{-1}e^{-umz} {\rm d} u$$
To me, this is the best of the forms.
You also notice, that $\Gamma(0,x)=E_1(x)$ (the exponential integral function). Similarly (pointed out by @MahmoundElawady), there is a generalized integro-exponential function that also includes a logarithm, so in that sense, you can write it down in "closed form" as:
$$I=E_1^1(mz)$$
A: $$I=\int_z^{\infty } \frac{\Gamma (0,m y)}{y} \, dy=\int_{mz}^{\infty }\frac{\Gamma (0,t )}{t} \, dt$$ Using a CAS,
$$I=\frac{\pi ^2}{12}+\frac{\gamma ^2}{2}+\frac{1}{2} \log ^2(m z)+\gamma  \log (m
   z)-m z \, _3F_3(1,1,1;2,2,2;-m z)$$
