7

Which method that will be effective for solving this integral?

Przemysław Scherwentke
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E.H.E
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  • Did you try partial fractions? – Vinícius Novelli Oct 28 '14 at 20:08
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    It doesn't converge. You can use Cauchy's Principal Value. – UserX Oct 28 '14 at 20:09
  • I think the partial fractions will not give me the solution – E.H.E Oct 28 '14 at 20:09
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    @Ehegh yeah of course it would not give you the solution, since the integral doesn't converge – Joshua Oct 28 '14 at 20:11
  • @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. – Brian M. Scott Oct 28 '14 at 20:12
  • @BrianM.Scott: how do you take the vertical asymptotes into account? Are you proposing the Cauchy Principal Value, because the integrals diverge in the standard sense. – robjohn Oct 28 '14 at 22:37
  • @robjohn: In the usual first-year calculus fashion: split the thing into six integrals, each improper at one end. Of course you’ll find that several of the integrals diverge, but that’s what you should find. In other words, the standard elementary methods work. – Brian M. Scott Oct 28 '14 at 22:41
  • @BrianM.Scott: okay, as long as you weren't saying that one could come up with a finite value for the integral without some sort of convention, such as the Cauchy Principal Value. – robjohn Oct 28 '14 at 22:43
  • @robjohn: No, of course not. (And I’m bemused by comments like Aaron’s above.) – Brian M. Scott Oct 28 '14 at 22:46
  • Please put the question in the question (as well as the title). On mobiles, the LaTeX of the title doesn't render. – Akiva Weinberger Nov 07 '14 at 01:16

3 Answers3

8

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}

Lucian
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mookid
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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.

robjohn
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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}

Felix Marin
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