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Could you explain to me, with details, how to compute this integral, find its principal value?

$$\int_0^{\infty} \frac{\sqrt{x}}{x^2-1}\mathrm dx$$

$f(z) =\frac{\sqrt{z}}{z^2-1} = \frac{z}{z^{1/2} (z^2-1)}$

It has singularities on the real line, so I need to integrate this function over a curve made of segment $[0, 1- \varepsilon] \cup $ semicircle with endpoints $1- \varepsilon, \ 1+ \varepsilon \cup$ segment $[1+ \varepsilon, R]$ and we link $R$ and $0$ with a big semicircle .

The integral over the big semicircle vanishes.

Integral over the small semicircle, centered in $1$ tends to $i \pi Res_1f$ as its radius tends to $0$.

Could you tell me how to calculate this put this together and find this integral?

Thank you.

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    $\begingroup$ Do you have to solve this integral a certain way? $\endgroup$
    – graydad
    Jan 25 '15 at 17:37
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    $\begingroup$ Are you asked to use contour integration? $\endgroup$ Jan 25 '15 at 17:43
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    $\begingroup$ The integral is divergent. So, I assume you want to calculate the principal value? $\endgroup$
    – mickep
    Jan 25 '15 at 17:43
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    $\begingroup$ Hint: For every $u\gt1$, use the change of variable $x\to1/x$ in the part $(0,1/u)$ of the integral on $(0,1/u)\cup(u,+\infty)$ to transform this into a computable integral on $(u,+\infty)$, then consider the limit $u\to1$. $\endgroup$
    – Did
    Jan 25 '15 at 17:54
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    $\begingroup$ In general, for $0<k<n$ we have $\displaystyle\int_0^\infty\frac{x^{k-1}}{1-x^n}dx=\frac\pi n~\cot\bigg(k~\frac\pi n\bigg)$. $\endgroup$
    – Lucian
    Jan 25 '15 at 21:44
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The integral as stated does not converge. On the other hand, its Cauchy principal value exists and may be computed using the residue theorem.

Consider the integral

$$\oint_C dz \frac{\sqrt{z}}{z^2-1} $$

where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$, with semicircular detours of radius $\epsilon$ into the upper half plane at $z=1$. The contour integral is then

$$\int_{\epsilon}^{1-\epsilon} dx \frac{\sqrt{x}}{x^2-1}+ i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\sqrt{1+\epsilon e^{i \phi}}}{\left (1+\epsilon e^{i \phi} \right )^2-1} \\ + \int_{1+\epsilon}^R dx \frac{\sqrt{x}}{x^2-1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\sqrt{R} e^{i \theta/2}}{R^2 e^{i 2 \theta}-1} \\ + \int_{R}^{1+\epsilon} dx \frac{e^{i \pi} \sqrt{x}}{x^2-1} + i \epsilon \int_{2\pi}^{\pi} d\phi \, e^{i \phi} \frac{\sqrt{e^{i 2 \pi}+\epsilon e^{i \phi}}}{\left (1+\epsilon e^{i \phi} \right )^2-1} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{e^{i \pi} \sqrt{x}}{x^2-1} + i \epsilon\int_{2 \pi}^0 d\phi \frac{\sqrt{\epsilon} e^{i \phi/2}}{\epsilon^2 e^{i 2 \phi}-1}$$

As $R \to \infty$, the magnitude of the fourth integral vanishes as $R^{-1/2}$. As $\epsilon \to 0$, each of the second integral contributes a factor of $-i \pi/2$, the sixth integral contributes a factor of $i \pi/2$ and the eighth integral vanishes as $\epsilon^{3/2}$. Also, the first, third, fifth, and seventh integrals add to the Cauchy principal value of an integral. Thus, the contour integral is equal to

$$2 PV \int_{0}^{\infty} dx \frac{\sqrt{x}}{x^2-1} $$

By the residue theorem, the contour integral is equal to $i 2 \pi$ times the residue of the pole $z=e^{i \pi}$ inside $C$. Thus,

$$PV \int_{0}^{\infty} dx \frac{\sqrt{x}}{x^2-1} = i \pi \frac{e^{i \pi/2}}{2 e^{i \pi}} = \frac{\pi}{2} $$

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  • $\begingroup$ The first integral is due to integrating along the segment $[\varepsilon, 1- \varepsilon]$ along the real line. Next, we integrate along the little semicircle till we touch the real line, then we integrate along the real line from $1- \varepsilon$ to $R$, then along the big semicircle, here $z = Re^{i \varphi}, \ 0 \le \varphi \le \pi$ $\endgroup$
    – Don
    Jan 25 '15 at 18:27
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    $\begingroup$ @Don: we double back along the real line after traversing the bug semicircle. Along the doubling back, $z=x e^{i 2 \pi}$. $\endgroup$
    – Ron Gordon
    Jan 25 '15 at 18:32
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    $\begingroup$ @Don: yes, but since we have a square root, we have to deal with different branches. Setting $z=x e^{i 2 \pi}$ is an expression of those branches. $\endgroup$
    – Ron Gordon
    Jan 25 '15 at 18:41
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    $\begingroup$ @Don: It's a big circle, not semicircle. Look up keyhole contour in wiki or on the site. $\endgroup$
    – Ron Gordon
    Jan 25 '15 at 18:58
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    $\begingroup$ @Don: yes, sort of. We have bumps above and below $z=1$. $\endgroup$
    – Ron Gordon
    Jan 25 '15 at 19:07
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Let $\delta_1>0,\delta_2<1$ real numbers close to zero and one, respectively, and I=$(\delta_1,\delta_2)\cup(\delta_2^{-1},\delta_1^{-1})$. We have: $$ \int_{I}\frac{\sqrt{x}}{x^2-1}\,dx = \int_{\delta_1}^{\delta_2}\frac{\sqrt{x}}{x^2-1}\,dx +\int_{\delta_1}^{\delta_2}\frac{\sqrt{1/x}}{1-x^2}\,dx = \int_{\delta_1}^{\delta_2}\frac{1}{x^{1/2}+x^{-1/2}}\frac{dx}{x}$$ and by setting $x=u^2$: $$ \int_{I}\frac{\sqrt{x}}{x^2-1}\,dx = 2\int_{\sqrt{\delta_1}}^{\sqrt{\delta_2}}\frac{du}{u^2+1}, $$ so, by letting $\delta_1\to 0,\delta_2\to 1$, we get that the principal value of the original integral is given by: $$ 2\arctan(1) = \color{red}{\frac{\pi}{2}}.$$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large a}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\mbox{P.V.}\int_{0}^{\infty}\frac{\root{x}}{x^{2} - 1}\,\dd x:\ {\large ?}}$.


We'll consider the $\ds{\root{z}}$-branch cut: $$ \root{z}=\verts{z}\exp\pars{\half\,{\rm Arg}\pars{z}\ic}\,,\qquad 0 < \,{\rm Arg}\pars{z} < 2\pi\,,\quad z \not= 0 $$ We'll perform an integration over a key-hole contour $\ds{\gamma}$ which takes care of the above branch-cut: $$ \oint_{\gamma}\frac{\root{z}}{z^{2} - 1}\,\dd z =2\pi\ic\bracks{\verts{\expo{\pi\ic}}^{1/2}\exp\pars{\half\,\pi\ic}\,\frac{1}{2\expo{\pi\ic}}}=\pi\tag{1} $$ The contribution from a 'big arc' of radius $\ds{R}$ goes as $\ds{R^{-1/2}}$ when $\ds{R \to \infty}$ while the contribution of a 'small arc' of radius $\ds{\epsilon}$, around the origin, goes as $\ds{\epsilon^{3/2}}$ when $\ds{\epsilon \to 0^{+}}$.

Then, \begin{align} \oint_{\gamma}\frac{\root{z}}{z^{2} - 1}\,\dd z &=\int_{0}^{\infty} \frac{\root{x}\expo{0\ic/2}}{\pars{x - 1 + \ic 0^{+}}\pars{x + 1}}\,\dd x +\int_{\infty}^{0} \frac{\root{x}\expo{2\pi\ic/2}}{\pars{x - 1 - \ic 0^{+}}\pars{x + 1}}\,\dd x \\[5mm]&=\int_{0}^{\infty}\root{x}\bracks{% \frac{1}{\pars{x - 1 + \ic 0^{+}}\pars{x + 1}} +\frac{1}{\pars{x - 1 - \ic 0^{+}}\pars{x + 1}}}\,\dd x \\[5mm]&=2\,\mbox{P.V.}\int_{0}^{\infty}\frac{\root{x}}{x^{2} - 1}\,\dd x\tag{2} \end{align} With $\pars{1}$ and $\pars{2}$: $$ \color{#66f}{\large\mbox{P.V.}\int_{0}^{\infty}\frac{\root{x}}{x^{2} - 1}\,\dd x} =\color{#66f}{\large\frac{\pi}{2}} $$

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Here is an approach. Recalling the Mellin transform

$$ F(s) = \int_{0}^{\infty} x^{s-1}f(x)dx.$$

We have the Mellin of our function $f(x)=\frac{1}{x^2-1}$ is given by

$$ F(s) = -\frac{\pi}{2} \cot(\pi s/2). $$

So our integral can be evaluated as

$$ I =\lim_{s\to 3/2}F(s) = \frac{\pi}{2} $$

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    $\begingroup$ Perhaps it seems odd that you are taking, as a given, a result more general than the result that needed proving. Also, I am not sure if the Mellin transform as you define it is a Cauchy principal value by default; the integral by which $F(s)$ is defined does not exist otherwise. $\endgroup$
    – Ron Gordon
    Jan 26 '15 at 20:35
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    $\begingroup$ BTW I understand your frustration with the downvotes. I took a look and see that your postings have been downvoted in a suspicious pattern lately. This happened to me last year and I did get some relief by contacting the Stack Exchange staff. $\endgroup$
    – Ron Gordon
    Jan 27 '15 at 13:08

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