Let $:\mathbb{R}\to (0,1)$ defined by $$F(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^xe^{-\frac{t^2}2}dt.$$
Note that $F$ is an increasing function, so $F^{-1}$ is well-defined. I have a conjecture that $\int_o^1F^{-1}(u)du$ is finite but I haven't a clue ho to check it? Any suggestion?
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I plugged this into mathematica and it told me that the answer was just $\mathbb{PI}/2$ but that can't be right be because f is a function of x and Pi over 2 does not depend on x. So I tried differentiating using the feynman integration rule. Introduce a new parameter $\mathbb{a}$ in the integral as so: $$\int_{-\infty}^{x} e^{-at^2} \mathrm{d}t$$. I know this looks like a different problem, but just wait. Lets differentiate twice with respect to a, and then once with respect to x, and then once with respect to t. It will not solve the problem but is good practice in differentiation. Ok now take another a derivative. You will get $$\int_{-\infty}^{x} t^2 e^{-at^2} \mathrm{d}t$$. Do a few more and add them all up in a series. $$\sum_{n=0}^{\infty} \int_{-\infty}^{x} t^{2n} e^{-at^2} \mathrm{d}t$$. This is called a poisson series and can be summed term by term to be: $$\int_{-\infty}^{x} \frac{e^{-at^2}}{1-t^{2}} \mathrm{d}t$$ This is easy. I plugged this into mathematica and it gave me the answer $\mathbb{PI}/4$. That looks right. mmhm.
$$\int_{0}^{1}F^{-1}(u)du=\int_{-\infty}^{\infty}zf(z)dz$$where I have used the substitution, $u=F(z)$ and used the fact that $dF(z)=f(z)dz$ where $f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. Then the result is just $0$.
In general, if $F(\cdot)$ is the cdf of a r.v. $X$, then what you seek is the mean $\mu$.