# An integral involving the inverse of f ( x ) = log &#x2061;<!-- ⁡ --> x

An integral involving the inverse of $f\left(x\right)=\mathrm{log}x-\mathrm{log}\mathrm{cos}x+x\mathrm{tan}x$
Let the function $f:\left(0,\phantom{\rule{thinmathspace}{0ex}}\frac{\pi }{2}\right)\to \mathbb{R}$ be defined as
$f\left(x\right)=\mathrm{log}x-\mathrm{log}\mathrm{cos}x+x\mathrm{tan}x.$
Let its inverse be denoted as ${f}^{\left(-1\right)}:\mathbb{R}\to \left(0,\phantom{\rule{thinmathspace}{0ex}}\frac{\pi }{2}\right)$, it satisfies

Consider the integral
$I={\int }_{0}^{1}z\phantom{\rule{thinmathspace}{0ex}}{e}^{z}\phantom{\rule{thinmathspace}{0ex}}{f}^{\left(-1\right)}\left(z\right)\phantom{\rule{thinmathspace}{0ex}}dz.$
Is it possible to evaluate the integral $I$ in a closed form?
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Antonio Dickerson
I know it has been some time now but we can transform this question into something more manageable. We'll just use the substitution $z=f\left(x\right)$ with $dz={f}^{\prime }\left(x\right)dx$. The bounds of integration change to the points x where $f\left(x\right)$ is respectively 0 or 1. Call these points ${x}_{0}$ and ${x}_{1}$.
Numerically they are
${x}_{0}=0.576412723\dots$
${x}_{1}=0.807626251\dots$
And so we have
$I={\int }_{{x}_{0}}^{{x}_{1}}f\left(x\right){e}^{f\left(x\right)}{f}^{\left(-1\right)}\left(f\left(x\right)\right){f}^{\prime }\left(x\right)dx={\int }_{{x}_{0}}^{{x}_{1}}x{f}^{\prime }\left(x\right)f\left(x\right){e}^{f\left(x\right)}dx$
Now we perform integration by parts with
$u=xf\left(x\right),\phantom{\rule{1em}{0ex}}du=\left(f\left(x\right)+x{f}^{\prime }\left(x\right)\right)dx$
$dv={f}^{\prime }\left(x\right){e}^{f\left(x\right)}dx,\phantom{\rule{1em}{0ex}}v={e}^{f\left(x\right)}$
and the integral becomes
$I={x}_{1}f\left({x}_{1}\right){e}^{f\left({x}_{1}\right)}-{x}_{0}f\left({x}_{0}\right){e}^{f\left({x}_{0}\right)}-{\int }_{{x}_{0}}^{{x}_{1}}f\left(x\right){e}^{f\left(x\right)}dx-{\int }_{{x}_{0}}^{{x}_{1}}x{f}^{\prime }\left(x\right){e}^{f\left(x\right)}dx$
Since $f\left({x}_{0}\right)=0$ and $f\left({x}_{1}\right)=1$ we have
$I={x}_{1}-{\int }_{{x}_{0}}^{{x}_{1}}f\left(x\right){e}^{f\left(x\right)}dx-{\int }_{{x}_{0}}^{{x}_{1}}x{f}^{\prime }\left(x\right){e}^{f\left(x\right)}dx$
Now we perform integration by parts on the latter integral with
$u=x,\phantom{\rule{1em}{0ex}}du=dx$
$dv={f}^{\prime }\left(x\right){e}^{f\left(x\right)}dx,\phantom{\rule{1em}{0ex}}v={e}^{f\left(x\right)}$
and we get
$I={x}_{1}-{x}_{1}{e}^{f\left({x}_{1}\right)}+{x}_{0}{e}^{f\left({x}_{0}\right)}-{\int }_{{x}_{0}}^{{x}_{1}}f\left(x\right){e}^{f\left(x\right)}dx+{\int }_{{x}_{0}}^{{x}_{1}}{e}^{f\left(x\right)}dx$
It turns out we can perform the integration in the first integral since
${e}^{f\left(x\right)}=\frac{x}{\mathrm{cos}x}{e}^{x\mathrm{tan}x}$
and
$\int {e}^{f\left(x\right)}dx={e}^{x\mathrm{tan}x}\mathrm{cos}x$
After cleaning up a bit
$I={x}_{0}+{e}^{{x}_{1}\mathrm{tan}{x}_{1}}\mathrm{cos}{x}_{1}-{e}^{{x}_{0}\mathrm{tan}{x}_{0}}\mathrm{cos}{x}_{0}-{\int }_{{x}_{0}}^{{x}_{1}}f\left(x\right){e}^{f\left(x\right)}dx$
and this is where I am getting stuck. Mathematica is having a problem with it too. On $\left[{x}_{0},{x}_{1}\right]$ we have
$f\left(x\right){e}^{f\left(x\right)}=\frac{x{e}^{x\mathrm{tan}\left(x\right)}}{\mathrm{cos}x}\left(\mathrm{ln}\left(\frac{x}{\mathrm{cos}x}\right)+x\mathrm{tan}x\right)$
We also have another form
$f\left(x\right){e}^{f\left(x\right)}=\left(\frac{{x}^{2}\mathrm{sin}x}{{\mathrm{cos}}^{2}x}+\frac{x}{\mathrm{cos}x}\mathrm{ln}\left(\frac{x}{\mathrm{cos}x}\right)\right){e}^{x\mathrm{tan}x}$
which suggest that we might have better luck looking at the following two integrals
$J={\int }_{{x}_{0}}^{{x}_{1}}\frac{{x}^{2}\mathrm{sin}x}{{\mathrm{cos}}^{2}x}{e}^{x\mathrm{tan}x}dx$
and
$K={\int }_{{x}_{0}}^{{x}_{1}}\mathrm{ln}\left(\frac{x}{\mathrm{cos}x}\right){e}^{x\mathrm{tan}x}dx$
The other problem perhaps, is that ${x}_{0}$ and ${x}_{1}$ are given numerically. I have no idea if one could find a closed for expression for them or even perhaps something like a series representation.
EDIT1: I'm not sure if this could be called a reduction but using the equation defining $f\left(x\right)$ we can say that
$\mathrm{cos}{x}_{1}{e}^{{x}_{1}\mathrm{tan}{x}_{1}}={x}_{1}{e}^{-1+2{x}_{1}\mathrm{tan}{x}_{1}}$
$\mathrm{cos}{x}_{1}0{e}^{{x}_{0}\mathrm{tan}{x}_{0}}={x}_{0}{e}^{2{x}_{0}\mathrm{tan}{x}_{0}}$