We have that $f:[1,\infty)\to\mathbb{R}$ is bounded and continuous, and we would like to show that $$\lim_{R\to\infty}\int_1^R\frac{f(t)}{t^2}\,\text{dt}$$ exists using the substitution rule.
Suppose $\lvert f\rvert\leq M$, $\forall x\in[1,\infty)$, then noting that continuous functions on compact sets are integrable, we can see
$$\Big{\lvert}\int_1^R\frac{f(t)}{t^2}\,\text{dt}\Big{\rvert}\leq M\Big{\lvert}\int_1^R\frac{1}{t^2}\,\text{dt}\Big{\rvert}$$
and then apply the substitution $\phi=-\frac{1}{t}$ so that $\text{d}\phi=\frac{1}{t^2}\text{d}t$ giving
$$M\Big{\lvert}\int_1^R\frac{1}{t^2}\,\text{dt}\Big{\rvert}= M\Big{\lvert}\int_{-1}^{-\frac{1}{R}}\text{dt}\Big{\rvert}= M(1-\frac{1}{R})$$
however I don't think this shows the result, since we are looking at the modulus of the integral and the integral could bounce in $[-M,M]$ as $R$ increases.
Without using the boundedness, $$\int_1^R\frac{f(t)}{t^2}\,\text{dt}= \int_{-1}^{-\frac{1}{R}}f\Big(-\frac{1}{\phi}\Big)\text{dt} $$ but I do not see how this could help.
I have tried some other similar substitutions but I can't seem to make progress with this. Any hints would be great!
For $ t\in (0,1] $, define
$$g(t)=f(\frac 1t)$$ $ f $ is bounded at $ [1,+\infty) \implies g $ is bounded at $ (0,1]$.
the substitution $ u=\frac 1t $, allows us to say the integrales $$I=\int_1^{+\infty}\frac{f(t)}{t^2}dt $$ and $$J=\int_0^1g(u)du $$ have the same nature.
Let $$M=\sup_{x\in(0,1]}|g(x)|$$ If we put, $ g(0)=M$, we prove easily that $ g $ is Riemann integrable at $ [0,1]$ and therefore
$$\lim_{r\to0^+}\int_r^{1}g(u)du\in \Bbb R$$ or $$\lim_{R\to+\infty}\int_1^R\frac{f(t)dt}{t^2}\in \Bbb R$$