I'm interested in the following integral:
$$\int_m^\infty\log(x+a) (1 + z(x-m))^{(-1-1/z)}dx$$
with $a>0$ and $m>0$ and $z\neq0$. All standard methods seem to fail here - is there anything to try? Was also playing around with differentiating under the integral sign (without introducing a new parameter but using the already existing $a$ - but got also stuck (as the differentiated integral $\frac{(1 + z(x-m))^{(-1-1/z)}}{x+a}$ does not become easier to solve).
Thanks for any clarification.
Define the function $\mathcal{I}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ via the improper integral
$$\mathcal{I}{\left(a,m,z\right)}:=\int_{m}^{\infty}\mathrm{d}x\,\left[1+z(x-m)\right]^{-1-1/z}\ln{\left(a+x\right)}.$$
Suppose $\left(a,m,z\right)\in\mathbb{R}_{>0}^{3}$, and set $a+m=:p>0\land\frac{1}{z}=:q>0$, followed by $\frac{q}{p}=:r>0$. After the appropriate shifting and scaling, we can obtain a much more compact expression for the integral $\mathcal{I}$ in terms of these auxiliary parameters:
$$\begin{align} \mathcal{I}{\left(a,m,z\right)} &=\int_{m}^{\infty}\mathrm{d}x\,\left[1+z(x-m)\right]^{-1-1/z}\ln{\left(a+x\right)}\\ &=\int_{0}^{\infty}\mathrm{d}y\,\left(1+zy\right)^{-1-1/z}\ln{\left(a+m+y\right)};~~~\small{\left[x=m+y\right]}\\ &=\int_{0}^{\infty}\mathrm{d}y\,\left(1+q^{-1}y\right)^{-1-q}\ln{\left(p+y\right)}\\ &=\int_{0}^{\infty}\mathrm{d}t\,q\left(1+t\right)^{-1-q}\ln{\left(p+qt\right)};~~~\small{\left[y=qt\right]}\\ &=\int_{0}^{\infty}\mathrm{d}t\,\frac{q}{\left(1+t\right)^{1+q}}\left[\ln{\left(p\right)}+\ln{\left(1+rt\right)}\right]\\ &=\ln{\left(p\right)}\int_{0}^{\infty}\mathrm{d}t\,\frac{q}{\left(1+t\right)^{1+q}}+\int_{0}^{\infty}\mathrm{d}t\,\frac{q}{\left(1+t\right)^{1+q}}\ln{\left(1+rt\right)}\\ &=\ln{\left(p\right)}\int_{0}^{\infty}\mathrm{d}t\,\frac{d}{dt}\left[\frac{(-1)}{\left(1+t\right)^{q}}\right]+\int_{0}^{\infty}\mathrm{d}t\,\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}\\ &=\ln{\left(p\right)}\left[\lim_{t\to\infty}\frac{(-1)}{\left(1+t\right)^{q}}-\lim_{t\to0}\frac{(-1)}{\left(1+t\right)^{q}}\right]+\int_{0}^{\infty}\mathrm{d}t\,\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}\\ &=\ln{\left(p\right)}\left[0-(-1)\right]+\int_{0}^{\infty}\mathrm{d}t\,\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}\\ &=\ln{\left(p\right)}+\int_{0}^{\infty}\mathrm{d}t\,\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}.\\ \end{align}$$
Given fixed but arbitrary $(q,r)\in\mathbb{R}_{>0}^{2}$, consider the following derivative w.r.t. $t$ on $\mathbb{R}_{>0}$:
$$\begin{align} \frac{d}{dt}\left[\frac{\ln{\left(1+rt\right)}}{\left(1+t\right)^{q}}\right] &=\frac{1}{\left(1+t\right)^{q}}\frac{d}{dt}\left[\ln{\left(1+rt\right)}\right]+\ln{\left(1+rt\right)}\frac{d}{dt}\left[\frac{1}{\left(1+t\right)^{q}}\right]\\ &=\frac{r}{\left(1+t\right)^{q}\left(1+rt\right)}-q\left(1+t\right)^{-1-q}\ln{\left(1+rt\right)}\\ &=\frac{1}{\left(1+t\right)^{q}\left(r^{-1}+t\right)}-\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}.\\ \end{align}$$
L'Hôpital's rule can be used to evaluate the following limit:
$$\begin{align} \lim_{t\to\infty}\frac{\ln{\left(1+rt\right)}}{\left(1+t\right)^{q}} &=\lim_{t\to\infty}\frac{\frac{d}{dt}\ln{\left(1+rt\right)}}{\frac{d}{dt}\left[\left(1+t\right)^{q}\right]}\\ &=\lim_{t\to\infty}\frac{\left(\frac{r}{1+rt}\right)}{q\left(1+t\right)^{q-1}}\\ &=\lim_{t\to\infty}\left[\frac{r\left(1+t\right)}{1+rt}\cdot\frac{1}{q\left(1+t\right)^{q}}\right]\\ &=\lim_{t\to\infty}\left[\frac{1+t}{r^{-1}+t}\cdot\frac{q^{-1}}{\left(1+t\right)^{q}}\right]\\ &=\left[\lim_{t\to\infty}\frac{1+t}{r^{-1}+t}\right]\cdot\left[\lim_{t\to\infty}\frac{q^{-1}}{\left(1+t\right)^{q}}\right]\\ &=1\cdot0\\ &=0.\\ \end{align}$$
We then find
$$\implies\frac{1}{\left(1+t\right)^{q}\left(r^{-1}+t\right)}-\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}=\frac{d}{dt}\left[\frac{\ln{\left(1+rt\right)}}{\left(1+t\right)^{q}}\right],$$
$$\implies\int_{0}^{\infty}\mathrm{d}t\,\left[\frac{1}{\left(1+t\right)^{q}\left(r^{-1}+t\right)}-\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}\right]=\int_{0}^{\infty}\mathrm{d}t\,\frac{d}{dt}\left[\frac{\ln{\left(1+rt\right)}}{\left(1+t\right)^{q}}\right],$$
$$\implies\int_{0}^{\infty}\mathrm{d}t\,\left[\frac{1}{\left(1+t\right)^{q}\left(r^{-1}+t\right)}-\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}\right]=\lim_{t\to\infty}\frac{\ln{\left(1+rt\right)}}{\left(1+t\right)^{q}}-\lim_{t\to0}\frac{\ln{\left(1+rt\right)}}{\left(1+t\right)^{q}},$$
$$\implies\int_{0}^{\infty}\mathrm{d}t\,\left[\frac{1}{\left(1+t\right)^{q}\left(r^{-1}+t\right)}-\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}\right]=0,$$
$$\implies\int_{0}^{\infty}\mathrm{d}t\,\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}=\int_{0}^{\infty}\mathrm{d}t\,\frac{1}{\left(1+t\right)^{q}\left(r^{-1}+t\right)}.$$
Recall that for any $\left(\alpha,\beta,\gamma,\xi\right)\in\mathbb{R}^{4}$ such that $0<\beta<\gamma\land\xi<1$, the following improper integral converges and can be expressed in terms of the Gauss hypergeometric function:
$$\int_{0}^{\infty}\mathrm{d}t\,\frac{t^{\gamma-\beta-1}\left(1+t\right)^{\alpha-\gamma}}{\left(1-\xi+t\right)^{\alpha}}=\operatorname{B}{\left(\beta,\gamma-\beta\right)}\,{_2F_1}{\left(\alpha,\beta;\gamma;\xi\right)}.$$
It follows from the integration formula above that
$$\int_{0}^{\infty}\mathrm{d}t\,\frac{\left(1+t\right)^{-q}}{\left(r^{-1}+t\right)}=q^{-1}{_2F_1}{\left(1,q;q+1;1-r^{-1}\right)};~~~\small{0<q\land0<r}.$$
Returning now to the evaluation of $\mathcal{I}$, we finally obtain
$$\begin{align} \mathcal{I}{\left(a,m,z\right)} &=\ln{\left(p\right)}+\int_{0}^{\infty}\mathrm{d}t\,\frac{q\ln{\left(1+rt\right)}}{\left(1+t\right)^{1+q}}\\ &=\ln{\left(p\right)}+\int_{0}^{\infty}\mathrm{d}t\,\frac{1}{\left(1+t\right)^{q}\left(r^{-1}+t\right)}\\ &=\ln{\left(p\right)}+q^{-1}{_2F_1}{\left(1,q;q+1;1-r^{-1}\right)}\\ &=\ln{\left(p\right)}+q^{-1}{_2F_1}{\left(1,q;q+1;1-\frac{p}{q}\right)}\\ &=\ln{\left(a+m\right)}+z\,{_2F_1}{\left(1,z^{-1};z^{-1}+1;1-(a+m)z\right)}.\\ \end{align}$$