Question Regarding Integration Within Summation 
It expresses an integral within a summation procedure.
 A: 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 $\ln\left(\frac{1+b}{1+a}\right)$.
A: $$\sum _{1}^{\infty}\log (\frac{k+a+1}{k+a})-\log(\frac{1+k+b}{k+b})$$
$$=\sum_1^\infty \log\frac{k+b(k+a+1)}{k+a(k+b+1)}$$
now sum it over 
and expanding the expression gives 
$$=\lim _{k \to \infty}\log \frac{(1+b)(2+a)}{(1+a)(2+b)}\frac {(2+b)(3+a)}{(2+a)(3+b)}.... \frac{(k+a+1)}{(k+b+1)}$$
as the consecutive expressions will cancel out in each iteration
$$=\lim _{k \to \infty} \log\frac{(1+b)(k+a+1)}{(1+a)(k+b+1)}$$
$$=\log\frac{1+b}{1+a}$$
