Approximate $\int_0^x e^{f^{-1}(t)} \; dt$ from an approximation of $f$

Question

I am wondering if it is possible to have an approximation of this integral

$\int_0^x e^{f^{-1}(t)} \; dt$

I have only an approximation of $f$: $f(\frac{i}{n}),\; i=0, \dots, n$?

Many thanks, Peter.

Answer 1

By the substitution $t=f(u)$ and integrating by parts, it can be shown that the integral equals

$$xe^{f^{-1}(x)}-\int^{f^{-1}(x)}_{f^{-1}(0)}f(u)e^udu$$

Assume $f$ is monotonically increasing (the decreasing case is similar). Since the domain of $f$ is $[0,1]$, we have $0\le f^{-1}\le 1$.

By assuming $f^{-1} (x)\approx 1, f^{-1}(0)\approx 0$, we can approximate the integral by Riemann sum $$\frac1n\sum^n_{i=1}f\left(\frac in\right)e^{i/n}$$

In conclusion, $$\int^x_0 e^{f^{-1}(t)}dt \approx xe^{f^{-1}(x)}-\frac1n\sum^n_{i=1}f\left(\frac in\right)e^{i/n} $$

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Error term bound

The error term $\epsilon :=\text{actual}-\text{approximation}$ can be bounded by $$\epsilon\le e\cdot f(1)(1-f^{-1}(x))+O\left(\frac1n\right)$$ $$\epsilon \ge xe^{f^{-1}(x)}(1-f^{-1}(x))+f^{-1}(0)f(0)-O\left(\frac 1n\right)$$