How to prove that the following function is monotonically increasing? $f(x) =  \frac{\sqrt{x}\int_a^1 e^{-xt} t^{\ b+1}~ \mathrm{d}t}{\int_a^1 e^{-xt} t^{\ b} ~\mathrm{d}t}:\ ]\ 0,+\infty\ [\ \to \mathbb{R} $
where  0 < a < 1   and  b > 0 .
By applying elementary rules of differentiation I can only prove that the function is monotonically increasing in $]\ 0,\frac{1}{1-a}\ ]$
But I need to prove that it is monotonically increasing in $]\ 0,+\infty\ [$
 A: Are you sure the function should have that property? I inserted $a = 1/10$ and $b=1$ which is in the allowed range. Then WA gives this result:
$$
F(x) = \frac{\sqrt{x}\int_{0.1}^1 e^{-xt}t^2dt}{\int_{0.1}^1e^{-xt}t dt} = \frac{e^{9x/10}(x^2 + 20x + 200) - 100(x^2 + 2x + 2)}{100\sqrt{x}(e^{9x/10}(x+10) - 10(x+1))}.
$$
Plotting that function from $0$ to $20$ shows that it is first increasing but then decreasing again.
A: $$f(x) =  \frac{\sqrt{x}\int_a^1 e^{-xt} t^{\ b+1}~ \mathrm{d}t}{\int_a^1 e^{-xt} t^{\ b} ~\mathrm{d}t}$$
Applying integration by parts
$$\int_a^1 e^{-xt} t^{\ b+1}dt = -\frac{1}{x}\int_a^1 e^{-xt} t^{\ b+1}dt + \frac{b+1}{x}\int_a^1 e^{-xt} t^{\ b}dt$$
Which means we can rewrite the functions as
$$f(x)=\frac{-\frac{1}{\sqrt{x}}\int_a^1 e^{-xt} t^{\ b+1}dt + \frac{b+1}{\sqrt{x}}\int_a^1 e^{-xt} t^{\ b}dt}{\int_a^1 e^{-xt} t^{\ b}dt}=\frac{f(x)}{-x}+\frac{b+1}{\sqrt{x}}$$
Now, we can solve for f(x):
$$f(x)=\frac{\sqrt{x}(b+1)}{x+1}$$
Now, it is much easier to reason about the function.
And as the previous answer mentioned, and as this graph shows, it is actually not monotonic for b=1
