<s>[INST] Question Regarding Integration Within Summation 
It expresses an integral within a summation procedure.
 [/INST] Compute everything from the inside out (one step at a time), don't look at the sum until you've considered the integral.
Observe that
$$
\int_{\frac{1}{k+b}}^{\frac{1}{k+a}}\frac{dx}{1+x}=\ln(1+x)|_{(k+b)^{-1}}^{(k+a)^{-1}}=\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right).
$$
By the properties of the logarithm, 
\begin{align*}
\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)&=\ln\left(\frac{k+1+a}{k+a}\right)-\ln\left(\frac{k+1+b}{k+b}\right)\\
&=\ln(k+1+a)-\ln(k+a)-\ln(k+1+b)+\ln(k+b).
\end{align*}
Now, let's consider the sum:
$$
\sum_{k=1}^\infty\ln(k+1+a)-\ln(k+a)-\ln(k+1+b)+\ln(k+b).
$$
This is a telescoping series, so the partial sum when $k=n$ is 
\begin{align*}
\sum_{k=1}^n\ln(k+1+a)-\ln(k+a)&-\ln(k+1+b)+\ln(k+b)\\
&=\ln(n+1+a)-\ln(n+1+b)-\ln(1+a)+\ln(1+b).
\end{align*}
Finally, by taking the limit as $n$ approaches infinity, we see that 
\begin{align*}
\lim_{n\rightarrow\infty}\ln(n+1+a)-\ln(n+1+b)
&=\lim_{n\rightarrow\infty}\ln\left(\frac{n+1+a}{n+1+b}\right)\\
&=\ln\left(\lim_{n\rightarrow\infty}\frac{n+1+a}{n+1+b}\right)\\
&=\ln(1)=0.
\end{align*}
Therefore, the limit is $\ln\left(\frac{1+b}{1+a}\right)$.
</s>