Let $f:[0,\infty) \rightarrow \mathbb R$ be a strictly positive, decreasing, differentiable function, such that $$f(0) = 1, \quad \lim_{x\rightarrow \infty} f(x) = 0$$ and $$\frac{1}{f(x)^2} = \frac{1}{f(x^2)} + 2x^2$$ If $\int_0^\infty f(x)\,dx$ exists, show that $$\int_0^\infty f(x^2) \,dx = \frac{1}{\sqrt2}\int_0^\infty f(x)\,dx$$
Integral for function of square
21
$\begingroup$
analysis
integration
-
0Is this homework? Did you try anything? Maybe a change of variables $u=x^2$? – 2012-03-21
-
0Hmm, it's not at all obvious to me how to proceed... from the information given I'm tempted to try something like writing both sides as functions of a second variable and doing some differentiation. +1, +star, and looking forward to seeing the answer. – 2012-03-21
-
1$f(x)=\frac{1}{2x}$ :) – 2012-03-21
-
0Martin, how did you come across this problem? I wouldn't be surprised if you had a means of constructing these types of identities... – 2012-03-22
-
0Thank you Antonio for your answer. It was just a simple observation that the function $\frac{1}{1 + x^2}$ satisfies the functional equation that you solved. – 2012-03-23