Integrals: can we take limits ‘partially’? Suppose I have an integral that converges properly
$$\int^b_a f(t)dt$$ where $f(t)$ is a complex analytic function.
($a,b\in\mathbb R, b>a$).
The integral is taken on a straight path.
I applied a substitution $u=\frac1{t-a+\frac1R}$, then we have the original integral equals to 
$$\int^R_{\frac1{b-a+\frac1R}}\frac{f(\frac1u+a-\frac1R)}{u^2}du$$
My question is, under what conditions is this true?

$$\int^b_a f(t)dt=\lim_{R\to\infty} \int^R_{\frac1{b-a+\frac1R}}\frac{f(\frac1u+a-\frac1R)}{u^2}du =\lim_{R\to\infty} \int^R_{\frac1{b-a}}\frac{f(\frac1u+a)}{u^2}du$$

I apologize for not being able to provide much context, as I have no idea how to tackle this problem.
 A: After the change of variables we already have for any $R \in (0,\infty)$
$$\int^R_{\frac1{b-a+\frac1R}}\frac{f(\frac1u+a-\frac1R)}{u^2}du = \int^b_a f(t)dt,$$
so
$$\lim_{R \to \infty}\int^R_{\frac1{b-a+\frac1R}}\frac{f(\frac1u+a-\frac1R)}{u^2}du = \int^b_a f(t)dt,$$
since for all $\epsilon > 0$, regardless of $R$,
$$\left|\int^R_{\frac1{b-a+\frac1R}}\frac{f(\frac1u+a-\frac1R)}{u^2}du - \int^b_a f(t)dt\right| = 0 < \epsilon$$
This is not entirely trivial in that the validity of change-of-variables has some conditions. However, integrability of $f$ and monotonicity of $u \mapsto 1/u +a - 1/R$ is sufficient.
For the second part apply the variable change $t = 1/u +a$ first to the integral over $[a + 1/R,b]$ to get
$$\int_{a + 1/R}^b f(t) \, dt = \int^R_{\frac1{b-a}}\frac{f(\frac1u+a)}{u^2}du.$$
We always have $\lim_{c \to a+}\int_c^b f(t) \, dt = \int_a^bf(t) \, dt$ if $f$ is properly integrable. Hence,
$$\int_a^b f(t) \, dt = \lim_{R \to \infty} \int_{a + 1/R}^b f(t) \, dt = \lim_{R \to \infty} \int^R_{\frac1{b-a}}\frac{f(\frac1u+a)}{u^2}du$$
