Monotonicity of $f(x)= \frac{ e^{\alpha x}-1}{\alpha x}$ where $\alpha \geq1 $ and $x >0$. There is this function I encountered when I was solving a problem and I am trying to study its monotonicity. So the function is $f(x)= \frac{ e^{\alpha x}-1}{\alpha x}$ where $\alpha \geq1 $ and $x >0$. 
I suspect that it is increasing for any $\alpha$ but getting the derivative I can not prove that it is positive. Can any propose any ideas? Thank you very much guys!
 A: Make a substution $y=\alpha x$ and examine  $g(y)=\frac{e^y - 1}{y}.$   Then 
$$
g'(y)= \frac{ye^y - e^y + 1}{y^2}.
$$
To show $g'(y) > 0$ for $y>0,$ it suffices to show that $(y-1)e^y > -1.$   Let $h(y) = (y-1)e^y.$  Then $h'(y)=y e^y,$ we have $h(0)=-1,$ $h'(0)= 0,$ and $h'(y)>0$ for $y>0,$  implying that $h(y)>-1$ for $y>0.$   Thus $g'(0)=0$ and $g'(y)>0$ for $y>0.$ 
A: $$f(x) := \frac{e^x-1}{x}=\int_0^1 e^{ux}du.$$
$e^{ux}, u>0$ increases with $x$, so does $f(x)$.
Moreover, by the same token, the first derivative 
$$\frac{ye^y - e^y + 1}{y^2}=f'(x) = \int_0^1 ue^{ux}du$$
also increases. In fact, arbitrary $n$'th derivative
$$\frac{d^nf(x)}{dx^n} = \int_0^1 u^ne^{ux}du$$
increases.
A: Note that $e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots$. For $t\gt 0$, subtract $1$, divide by $t$. We get $1+\frac{t}{2!}+\frac{t^2}{3!}+\cdots$.
This is a series with positive coefficients, so is increasing for $t\gt 0$. 
A: The derivative of the function $f(x)$ is 
$$
f'(x)=\frac{\alpha xe^{\alpha x}-e^{\alpha x}+1}{x^2}.
$$
Since the denominator is always positive, it suffices to study sign of the numerator.  In other words,
$$
(\alpha x-1)e^{\alpha x}+1=0
$$
Using the substitution $y=\alpha x-1$, we have 
$$
ye^{y+1}+1=0.
$$
In other words, for $y>-1$, we would like to find if 
$$
ye^y=-\frac{1}{e} 
$$
has a solution.  When $y=-1$, there is a solution, but $y>-1$.  
Now, consider $ye^y$, the derivative of this function is $(1+y)e^y$.  Since $y>-1$, this derivative is always positive (because both terms are positive).  Therefore, $ye^y=-\frac{1}{e}$ when $y=-1$ and the LHS is increasing for all $y>-1$.  Therefore, there is no solution to $ye^y=-\frac{1}{e}$ in the interval $y>-1$ and the equality can never hold.  Moreover, for all $y>-1$, $ye^y>-\frac{1}{e}$, so the original function is increasing.
