7
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Which method that will be effective for solving this integral?

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  • $\begingroup$ Did you try partial fractions? $\endgroup$ – Vinícius Novelli Oct 28 '14 at 20:08
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    $\begingroup$ It doesn't converge. You can use Cauchy's Principal Value. $\endgroup$ – UserX Oct 28 '14 at 20:09
  • $\begingroup$ I think the partial fractions will not give me the solution $\endgroup$ – E.H.E Oct 28 '14 at 20:09
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    $\begingroup$ @Ehegh yeah of course it would not give you the solution, since the integral doesn't converge $\endgroup$ – Joshua Oct 28 '14 at 20:11
  • $\begingroup$ @Ehegh: Partial fractions is the standard elementary solution and will work. Of course, you have to realize that the function has vertical asymptotes at $\frac14$ and $\frac34$ and take these into account. $\endgroup$ – Brian M. Scott Oct 28 '14 at 20:12
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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

$\hspace{4.5cm}$enter image description here

$$ \small\color{#00A000}{iR[1,0]}\cup\color{#C00000}{\left[0,\frac14-\frac1R\right]}\cup\color{#00A000}{\frac14+\frac1Re^{i\pi[1,0]}}\cup\color{#C00000}{\left[\frac14+\frac1R,\frac34-\frac1R\right]}\cup\color{#00A000}{\frac34+\frac1Re^{i\pi[1,0]}}\cup\color{#C00000}{\left[\frac34+\frac1R,R\right]}\cup \color{#0000FF}{Re^{i\pi\left[0,\frac12\right]}} $$ 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.

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  • $\begingroup$ nice picture!${}$ $\endgroup$ – mookid Oct 28 '14 at 22:46
8
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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}

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  • $\begingroup$ Neither of the integrals converge since each contains a singularity; one at $x=\frac34$ and one at $x=\frac14$. $\endgroup$ – robjohn Oct 28 '14 at 21:39
  • $\begingroup$ @robjohn I fixed the typo, thanks. $\endgroup$ – mookid Oct 28 '14 at 21:53
  • $\begingroup$ What happens as $a\to\frac34^+$? $-\frac18\log\left(\frac{4a-3}{4a-1}\right)\to+\infty$ $\endgroup$ – robjohn Oct 28 '14 at 22:13
  • $\begingroup$ Look at the beginning of my post: "The integral does not converge because of the singularities. To compute the integral between $a$ and $\infty$ for $a>\frac 34$..." $\endgroup$ – mookid Oct 28 '14 at 22:42
  • $\begingroup$ Ah, I missed the change there. $\endgroup$ – robjohn Oct 28 '14 at 22:44
1
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

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