I've seen that $I$ satisfies that $$\lim_{x\to 0^{+}}I(x)=0$$and also that$$\lim_{x\to+\infty}I(x)=\lim_{x\to\infty}\frac{1}{\sqrt{e^{x}}}\int_{0}^{x}\frac{dt}{\sqrt{1-\frac{e^{t}}{e^{x}}}}=0,$$ so by Rolle's theorem there must be at least one maximum in $(0,+\infty)$. And I have to prove that there is only one maximum, I just don't seem to be able to do it because when I use Leibniz's integral rule I get that $$I'(x)=\frac{1}{\sqrt{e^{x}-e^{x}}}+\int_{0}^{x}\frac{e^{x}}{2\sqrt{\left(e^{x}-e^{t}\right)^{3}}}dt$$ which explodes because of the first term. I don't know how to get further than this step.
Although it's better to have a mathematical proof, it would also be helpful to be able to plot the function on MatLab, it just keeps giving me dimension errors somehow. All help will be welcome.
Best Answer
$$I=\int \frac{dt}{\sqrt{e^{x}-e^{t}}}=\int \frac{dt}{e^{x/2}\sqrt{1-e^{(t-x)}}}$$ $$t=x+\log(u) \implies dt=\frac {du}u\implies I=e^{x/2}\int\frac {du}{u \sqrt{1-u} }$$ $$u=1-v^2 \implies \int\frac {du}{u \sqrt{1-u} }=2\int \frac {dv}{1-v^2}=-2 \tanh ^{-1}(v) $$
Back to $x$ and using the bounds leads to $$J(x)=\int_0^x \frac{dt}{\sqrt{e^{x}-e^{t}}}=2 e^{-x/2} \tanh ^{-1}\left(e^{-x/2} \sqrt{e^x-1}\right)$$ $$J'(x)=\frac{1}{\sqrt{e^x-1}}-e^{-x/2} \tanh ^{-1}\left(e^{-x/2} \sqrt{e^x-1}\right)$$
By inspection or graphing, the solution is just above $x=1$ and Newton methods gives the folloxing iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.0000000 \\ 1 & 1.1597496 \\ 2 & 1.1862399 \\ 3 & 1.1868419 \\ 4 & 1.1868422 \end{array} \right)$$ and, at this point, $J''(x)=-0.47692$ confirms that this is a maximum.