How do I show the function $I:\mathbb{R}^{+}\rightarrow\mathbb{R}$ defined by $I(x)=\int_{0}^{x} \frac{dt}{\sqrt{e^{x}-e^{t}}}$ has a unique maximum?
Solution 1:
$$I=\int \frac{dt}{\sqrt{e^{x}-e^{t}}}=\int \frac{dt}{e^{x/2}\sqrt{1-e^{(t-x)}}}$$ $$t=x+\log(u) \implies dt=\frac {du}u\implies I=e^{x/2}\int\frac {du}{u \sqrt{1-u} }$$ $$u=1-v^2 \implies \int\frac {du}{u \sqrt{1-u} }=2\int \frac {dv}{1-v^2}=-2 \tanh ^{-1}(v) $$
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.