Finding the integral $\int_{0}^{\infty }\frac{1}{(4x-3)(4x-1)}\,dx$. Which method that will be effective for solving this integral?
 A: The integral does not converge because of the singularities.
To compute the integral between $a$ and $\infty$ for $a>\frac 34$, you can use the following method:
\begin{align}
\int_{a}^{\infty }\frac{1}{(4x-3)(4x-1)} dx
&= \lim_{N\to\infty} \int_{a}^N \frac{1}{(4x-3)(4x-1)} dx
\end{align}
Now use the fact that $$
\frac{1}{(4x-3)(4x-1)} =
\frac 12 \left(
\frac 1{4x-3} - \frac 1{4x-1}
\right)
$$
so that
\begin{align}
\int_{a}^N \frac{dx}{(4x-3)(4x-1)} 
&= \frac 12 \left(
\int_{a}^N \frac {dx}{4x-3} dx - \int_a^N \frac {dx}{4x-1} dx
\right) 
\\&= \frac 18\left(
\log \frac{|4N-3|}{|4a-3|} - \log \frac{|4N-1|}{|4a-1|}
\right)
\\&= \frac 18\left(
\log \frac{4N-3}{4N-1} - \log \frac{4a-3}{4a-1}\right)
 \to -\frac18 \log \frac{4a-3}{4a-1}
\end{align}
A: As stated, the integral does not converge. It contains two singular points at which the integral does not converge: $x=\frac14$ and $x=\frac34$.
However, we can apply the Cauchy Principal Value. One way to compute this is using contour integration. We can use the contour
$\hspace{4.5cm}$
$$
\small\color{#00A000}{iR[1,0]}\cup\color{#C00000}{\left[0,\frac14-\frac1R\right]}\cup\color{#00A000}{\frac14+\frac1Re^{i\pi[1,0]}}\cup\color{#C00000}{\left[\frac14+\frac1R,\frac34-\frac1R\right]}\cup\color{#00A000}{\frac34+\frac1Re^{i\pi[1,0]}}\cup\color{#C00000}{\left[\frac34+\frac1R,R\right]}\cup \color{#0000FF}{Re^{i\pi\left[0,\frac12\right]}}
$$
as $R\to\infty$ to get
$$
\begin{align}
\mathrm{PV}\int_0^\infty\frac{\mathrm{d}x}{(4x-3)(4x-1)}
&=\pi i\left(\operatorname*{Res}_{z=1/4}\frac1{(4z-3)(4z-1)}+\operatorname*{Res}_{z=3/4}\frac1{(4z-3)(4z-1)}\right)\\
&+\int_0^\infty\frac1{(4ix-3)(4ix-1)}i\,\mathrm{d}x\\
&=\pi i\left(-\frac18+\frac18\right)+\frac18\left[\log\left(\frac{4ix-3}{4ix-1}\right)\right]_0^\infty\\
&=-\frac{\log(3)}{8}
\end{align}
$$
The contour contains no singularities, so the integral over the whole contour is $0$. The red pieces are the Cauchy Principal Value. The green pieces are the negative of the residues and integral above. The blue piece vanishes as $R\to\infty$.
In this case, the Cauchy Principal Value equals what one would get if one naively uses the formulas gotten by ignoring the singularities, but this is not always the case.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{0 < a < \Lambda}$:

\begin{align}&\color{#c00000}{\pp\int_{0}^{\Lambda}{\dd x \over x - a}}
=\lim_{\epsilon\ \to\ 0^{+}}\pars{\int_{0}^{a - \epsilon}{\dd x \over x - a}
+\int_{a + \epsilon}^{\Lambda}{\dd x \over x - a}}
=\lim_{\epsilon\ \to\ 0^{+}}
\ln\pars{\verts{-\epsilon\bracks{\Lambda - a} \over -a\epsilon}}
\\[5mm]&=\ln\pars{\Lambda - a \over a}
\end{align}

\begin{align}&\color{#66f}{\large%
\pp\int_{0}^{\infty}{\dd x \over \pars{4x - 3}\pars{4x - 1}}}
={1 \over 8}\,\lim_{\Lambda\ \to\ \infty}\bracks{
\pp\int_{0}^{\Lambda}{\dd x \over x - 3/4}
-\pp\int_{0}^{\Lambda}{\dd x \over x - 1/4}}
\\[5mm]&={1 \over 8}\,\lim_{\Lambda\ \to\ \infty}
\bracks{\ln\pars{\Lambda - 3/4 \over 3/4} - \ln\pars{\Lambda - 1/4 \over 1/4}}
={1 \over 8}\,\lim_{\Lambda\ \to\ \infty}
\ln\pars{{1 \over 3}\,{\Lambda - 3/4 \over \Lambda - 1/4}}
\\[5mm]&=\color{#66f}{\large -\,{1 \over 8}\,\ln\pars{3}} \approx {\tt -0.1373}
\end{align}
