# Anomalous integral equation

I'm trying to solve the following equation: $$\int_0^{f(x)}f(t)dt=g(x)$$ Differentiating under integral I obtain: $$f[f(x)]\frac{d}{dx}f(x)=\frac{d}{dx}g(x)$$ I know the function $g(x)$. Is there a simple way to find the function $f(x)$? Is it possible to find it without the rule of the derivative under integral? Thanks in advance.

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I do not get how do you do "diiferentaiting under the integral sign" - there is no parameter dependent integrand here... – Dirk Mar 12 '12 at 14:01
Did you use this rule: en.wikipedia.org/wiki/Leibniz_integral_rule – Dirk Mar 12 '12 at 14:03

Let us consider the following change of variable in the integral

$$z=f^{-1}(t)$$

then

$$dz=\frac{d}{dt}f^{-1}(t)dt.$$

One has

$$\int_{f^{-1}(0)}^x f[f(z)]\left[\frac{d}{dt}f^{-1}(t)\right]^{-1}_{t=f(z)}dz=\int_{f^{-1}(0)}^x f[f(z)]\frac{d}{dz}f(z)dz=g(x)$$

that is equivalent to the differential equation you give. But this permits to use integration by parts giving

$$f[f(x)]f(x)-\int_{f^{-1}(0)}^x f(z)\frac{df[f(z)]}{dz}\frac{df(z)}{dz}dz=g(x)$$

and the procedure can be repeated to give

$$f[f(x)]f(x)-[f(x)]^2\frac{df[f(x)]}{dx}+\int_{f^{-1}(0)}^x f(z)\frac{d}{dz}\left\{f(z)\frac{df[f(z)]}{dz}\frac{df(z)}{dz}\right\}dz=g(x).$$

Taking this as an asymptotic series in powers of $f$ or just neglecting derivatives of the function $f(x)$ you will get a first approximate solution by inverting the equation

$$f[f(x)]f(x)\approx g(x)$$

that is a functional equation.

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