An integrable function $f: \mathbb{R} \to \mathbb{R}$ such that $a f(a) + b f(b) = 0$ when $ab = 1$.

Question

Consider an integrable function $f: \mathbb{R} \to \mathbb{R}$ such that $a f(a) + b f(b) = 0$ when $ab = 1$.

Find the value of the following integration: $$\int_0^{+\infty}f(x)dx$$

I know how to do integration but how can I find the function $f(x)$ under the given condition .

Answer 1

Split the integral into two: $$\int_{0}^{+\infty}f(x)dx = \int_{0}^1f(x)dx + \int_{1}^\infty f(x)dx$$ Perform the variable substitution $x = \frac{1}{a}$ in the second one, then use the expression from the problem statement: $$\frac{1}{a^2}f\left(\frac{1}{a}\right) = -f(a)$$