Analyze the variation of this function $f(x)=x \exp[a+b(x-1)] \, E_1[a+b (x-1)]$ w.r.t. $x$ Please, I need to analyse the variation of the following function w.r.t. $x$ :
$f(x)=x \exp[a+b(x-1)] \, E_1[a+b (x-1)]$, where  $E_1[a+b  (x-1)]$ is the exponential integral, $b>a$, $a>0$, $b>0$ and $x \ge 1$. Note that I don't have a specific values for $a$ and $b$.
The first derivative $\frac{df(x)}{dx}=$
$\exp[a+b(x-1)]\, E_1[a+b (x-1)]+ xb \exp[a+b(x-1)] \, E_1[a+b (x-1)]- \frac{xb}{a+b(x-1)}$.
I am stuck here.
Any help is welcome! 
 A: What follows is not a complete answer.
Perhaps it is a good idea to let $c=b-a>0$ and $y = a + b(x-1) > a$, and then study the variations of $g(y) = (y+c)e^y E_1(y)$, which looks much simpler -- check that you have $f(x) = g(y) / b$. Since you have exponentials and products, you might also want to take the logarithmic derivative instead :
$$ \frac{g'(y)}{g(y)} = \frac1{y+c} + 1 + \frac{E_1'(y)}{E_1(y)}. $$
You want to know the values of $y>a$ where this vanishes, depending on $c$. But actually there is only one occurrence of $c$ in that equation : so it might be simpler to solve for $c$ first in terms of $y$, and then try to invert. Solving for $c$ yields
$$ c = K(y) := \frac1{-E_1'/E_1(y) - 1} - y. $$
Plotting that function of Wolfram Alpha or Maple, you guess that it is strictly increasing -- if that was true, then for each $y>0$, there would only be one solution $c > 0$. But if that was true, by taking reciprocals (since the function $K(y)$ is assumed to be strictly increasing), you would get a single root $y>0$ for each value of $c>0$. Then you would have to check that such a root gives you back a value of $x>1$ : this is done by checking that $y>a$, or in other terms, $c>K(a)$.
Then there remains to analyze $K(y)$. That function does look like it behaves nicely but I don't see a clever way to proceed. How did that question come up ?
