Let $X$ be log-normally distributed. That is, $X=e^Y$ where $\ln Y \sim N(\mu,\sigma^2)$. The CDF of $X$
$$F(x)=\Phi\left(\frac{\ln x -\mu}{\sigma}\right),$$
where $\Phi$ is the CDF of a normal distribution which is
$$ F_Y(y)=\Phi(y) =\frac{1}{\sigma\sqrt{2\pi}}\int^y_{-\infty}\exp \left(-\frac{1}{2}\left(\frac{u-\mu}{\sigma}\right)^2\right)du.$$
I am trying to solve the Lorenz curve of $X$. The definition of Lorenz curve is
$$L(p)=\frac{\int^{p}_{0}F^{-1}(t)dt}{\mu}=\frac{\int^{p}_{0}F^{-1}(t)dt}{\int^{1}_{0}F^{-1}(t)dt},$$
where $F^{-1}(t)$ is the inverse of the CDF of $X$.
I have already solved the $F^{-1}(t)$ which is $\exp(\sigma\Phi^{-1}(t)+\mu)$. The problem I have is that I don't know how to integrate this. I tried substitution
$$\int F^{-1}(t)dt=\int \exp(\sigma\Phi^{-1}(t)+\mu)dt)=\int e^udu,$$
where $u$ is $\sigma\Phi^{-1}(t)+\mu$. However, I don't know how to get a derivative of it to get $du$.
Let $$I(t)=\int\exp\left(\sigma\Phi^{-1}(t)+\mu\right)\,\mathrm dt.$$
Substitution $$t = \Phi(z) = \dfrac12\left(1+\operatorname{erf}\left(\dfrac z{\sqrt2}\right)\right)$$
allows to write $$I(\Phi(z)) = \int\,e^{\sigma z+\mu}\cdot\dfrac1{\sqrt{2\pi}}e^{-\,^1/_2\,z^2}\,\mathrm dz = \dfrac12e^{\,^1/_2\,\sigma^2+\mu}\,\operatorname{erf}\left(\dfrac{z-\sigma}{\sqrt2}\right) + \mathrm{constant}$$ (see also Wolfram Alpha calculations),
$$I(t) = e^{\,^1/_2\,\sigma^2+\mu}\,\Phi\left(\Phi^{-1}(t)-\sigma\right) + \mathrm{constant}.$$