If $ \int_1^\infty f(x)dx $ converges then $ \int_0^1 \frac{1}{x^2} f(1/x)dx $ converges

Question

Prove/Disprove ( here $ f $ is Riemann Integrable on $[0,\infty) $ ): If $ \int_1^\infty f(x)dx $ converges then $ \int_0^1 \frac{1}{x^2} f(1/x)dx $ converges

Attempt:
Maybe it's a disprove? I couldn't find a counter example. I tried the following functions $ f(x) = \exp(-x - 1/x) $, $ f(x) = \frac{1}{x^p} $ where $ p $ is some real number , $ f(x) = \frac{\cos x}{1+x} $ , $ f(x) = \frac{\ln(1+x)}{x^2} $, but everyone seems to satisfy the theorem above. If it's a true statement, how can one approach it? , if it's a false statement then how can I think of a function that satisfies the above without randomly sampling some functions, that satisfy $ \int_1^\infty f(x)dx $ ?

Answer 1

The change of variable $y=\frac 1x$ converts the second integral to the first, so the assertion is true.