I'm reading a paper and I can't understand this variable change $$ \int_{0}^{\infty}f\left(\frac{t}{C\lambda^{x}}\right)dx = \frac{1}{\log{\lambda}}\int_{0}^{\frac{t}{C\lambda}}\frac{f(y)}{y}dy $$ I would be very grateful if someone can help me in any way.
edit: $\lambda > 1, t>0$ and $C> 0$.
I'm not sure whether your equation is entirely correct. In fact, for $C = t = 1, \lambda = 2$ and $f = \mathrm{id}$ WolframAlpha calculates the right hand side as approximately $1.4$ but the left hand side as approximately $0.7$.
A only slightly different result is obtained by using substitution with the inverse function of $\varphi: x \mapsto {t \over {C \lambda^x}}$ which is $$\varphi^{-1}: y \mapsto - \frac{\log(yC) - \log(t)}{\log \lambda}$$ with $$(\varphi^{-1})': y \mapsto - \frac{1}{\log \lambda} \frac{1}{yC} \cdot C = - \frac{1}{y \cdot \log \lambda}.$$ Then the substitution formula gives us $$ \int_0^\infty f\left(\frac{t}{C \lambda^x}\right) dx = \int_{\varphi(0)}^{\varphi(\infty)} f(y) \cdot (\varphi^{-1})'(y) dy = \int_0^{\frac{t}{C}} \frac{f(y)}{y \log \lambda} dy, $$ where I assumed that $| \lambda | > 1$ which gives $\lim_{x \rightarrow \infty} \varphi(x) = \lim_{x \rightarrow \infty} \frac{t}{C \lambda^x} = 0$.