I need your help with this integral:

$$\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}$$

where $a>0$. I have tried some complex integration methods, but none seems adequate for this particular one.

Is there a specific method for this kind of integrals? What contour should I use?


$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} &\color{#f00}{% \int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x^{2} + a^{2}}^{2}}\,\dd x} \,\,\,\stackrel{x\ \to\ x^{1/2}}{=}\,\,\, {1 \over 4}\int_{0}^{\infty} {x^{-3/4}\,\ln\pars{x} \over \pars{x + a^{2}}^{2}}\,\dd x \\[5mm] = &\ -\,{1 \over 4}\,\partiald{}{\pars{a^{2}}}\int_{0}^{\infty} {x^{-3/4}\,\ln\pars{x} \over x + a^{2}}\,\dd = -\,{1 \over 8\verts{a}}\,\partiald{}{\verts{a}}\int_{0}^{\infty} {x^{-3/4}\,\ln\pars{x} \over x + a^{2}}\,\dd x \\[5mm] = &\ -\,{1 \over 8\verts{a}}\,\partiald{}{\verts{a}}\bracks{% \lim_{\mu \to -3/4}\,\,\partiald{}{\mu} \int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x}\tag{1} \end{align}

With the branch-cut $\ds{z^{\mu} = \verts{z}^{\mu}\exp\pars{\ic\,\mathrm{arg}\pars{z}\mu}\,,\ 0 < \mathrm{arg}\pars{z} < 2\pi\,,\ z \not = 0}$, the integral is performed along a key-hole contour. Namely, \begin{align} 2\pi\ic\,\verts{a}^{2\mu}\exp\pars{\ic\pi\mu} & = \int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x + \int_{\infty}^{0}{x^{\mu}\exp\pars{2\pi\mu\ic} \over x + a^{2}}\,\dd x \\[3mm] & = -\exp\pars{\ic\pi\mu}\bracks{2\ic\sin\pars{\pi\mu}} \int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x \\[5mm] \imp\ \int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x & = -\pi\,\verts{a}^{2\mu}\csc\pars{\pi\mu} \end{align}

Plug this result in $\pars{1}$: $$ \color{#f00}{% \int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x^{2} + a^{2}}^{2}}\,\dd x} = \color{#f00}{{\root{2} \over 16}\,\pi\,{% 6\ln\pars{\verts{a}} - 3\pi - 4 \over \verts{a}^{7/2}}} $$

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  • 2
    $\begingroup$ I will study this. Upvoted. (+1). The page is starting to look interesting. $\endgroup$ – Marko Riedel Jun 4 '16 at 22:59
  • $\begingroup$ Great answer, thank you! $\endgroup$ – guest_user Jun 5 '16 at 13:28

We integrate $$f(z) = \frac{\log z}{\sqrt{z}(z^2+a^2)^2}$$ along a keyhole contour with the branch cut of the logarithm on the positive real axis and its argument between $0$ and $2\pi.$ There is a small circle of radius $\epsilon$ enclosing the origin and the large circle has radius $R.$

Along the straight segment on the positive real axis we get in the limit

$$I = \int_0^\infty \frac{\log x}{\sqrt{x}(x^2+a^2)^2} \; dx.$$

Just below the real axis we get with $\sqrt{z} = \exp(1/2\log z)$

$$- \int_0^\infty \frac{\log x + 2\pi i}{\exp(\pi i)\sqrt{x}(x^2+a^2)^2} \; dx = I + 2\pi i \int_0^\infty \frac{1}{\sqrt{x}(x^2+a^2)^2} \; dx = I + 2\pi i J.$$

Now we need to invoke a recursive step and compute $J$ using the same keyhole contour. We get $J$ on the positive real axis and just below we get with $$g(z) = \frac{1}{\sqrt{z}(z^2+a^2)^2}$$

the integral $$- \int_0^\infty \frac{1}{\exp(\pi i)\sqrt{x}(x^2+a^2)^2} \; dx = J.$$

It follows that

$$2J = 2\pi i (\mathrm{Res}_{z=ai} g(z)+\mathrm{Res}_{z=-ai} g(z)).$$

We get for the two residues

$$\lim_{z=\pm ai} \left(\frac{\exp(-1/2\log(z))}{(z\pm ai)^2}\right)' \\ = \lim_{z=\pm ai} \left(\frac{-1/2\exp(-1/2\log z)/z}{(z\pm ai)^2} - 2\frac{\exp(-1/2\log z)}{(z\pm ai)^3}\right)$$

We obtain for the first residue

$$-\frac{3}{16} (1+i)\sqrt{2} a^{-7/2}$$

and for the second one

$$\frac{3}{16} (1-i)\sqrt{2} a^{-7/2}.$$

It follows that

$$J = \pi i \times -\frac{3}{8} i \sqrt{2} a^{-7/2} = \frac{3}{8} \pi \sqrt{2} a^{-7/2}.$$

Returning to the main computation we thus have

$$2I+2\pi i J = 2\pi i(\mathrm{Res}_{z=ai} f(z)+\mathrm{Res}_{z=-ai} f(z)).$$

This time we obtain for the two residues

$$\lim_{z=\pm ai} \left(\frac{\exp(-1/2\log(z))\log z}{(z\pm ai)^2}\right)' \\ = \lim_{z=\pm ai} \left(\frac{-1/2\log z \times \exp(-1/2\log z)/z + \exp(-1/2\log(z))/z}{(z\pm ai)^2} \\ - 2\frac{\exp(-1/2\log z)\log z}{(z\pm ai)^3}\right)$$

We get for the first residue

$$-\frac{1}{32} (1+i)\sqrt{2} (3i\pi + 6\log a - 4) a^{-7/2}$$

and for the second one

$$\frac{1}{32} (1-i)\sqrt{2} (9i\pi + 6\log a - 4) a^{-7/2}$$

Adding these two yields

$$-\frac{1}{16} i\sqrt{2} (6i\pi + 6\log a - 3\pi - 4) a^{-7/2}.$$

Multiply by $\pi i$ and subtract $\pi i J$ to get

$$\frac{1}{16}\pi \sqrt{2} (6i\pi + 6\log a - 3\pi - 4) a^{-7/2} - \frac{3}{8} \pi \sqrt{2} \times \pi i\times a^{-7/2}$$

for a final answer of

$$\bbox[5px,border:2px solid #00A000] {\frac{1}{16}\pi \sqrt{2} (6\log a - 3\pi - 4) a^{-7/2}.}$$

The bounds here are mostly trivial, e.g. when computing $I$ we get by ML for the large circle of radius $R$

$$\lim_{R\rightarrow\infty} 2\pi R \times \frac{|\log R+2\pi i|}{\sqrt{R}(R^2-a^2)^2} = \lim_{R\rightarrow\infty} 2\pi R \times \frac{\sqrt{\log^2 R+4\pi^2}}{\sqrt{R}(R^2-a^2)^2} = 0$$

and for the small circle

$$\lim_{\epsilon\rightarrow 0} 2\pi \epsilon \times \frac{|\log \epsilon+2\pi i|}{\sqrt{\epsilon}(a^2-\epsilon^2)^2} = \lim_{\epsilon\rightarrow 0} 2\pi \times \frac{\sqrt\epsilon \sqrt{\log^2 \epsilon + 4\pi^2}}{(a^2-\epsilon^2)^2} = 0.$$

Remark. We have to use the same branch of the logarithm throughout e.g. $\log(-i) = 3i\pi/2$ as opposed to $\log(-i) = -i\pi/2.$

Addendum 05 Jun 2016. It appears the computation of $J$ is not necessary. We can just take the real part of the contributions from the poles of $f(z)$ because we know $I$ and $J$ are real numbers.

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  • $\begingroup$ Great answer! about the bound for the small circle, did you mean $\epsilon \rightarrow 0$? $\endgroup$ – Yagger Sep 7 '19 at 18:26
  • $\begingroup$ Thanks, corrected. $\endgroup$ – Marko Riedel Sep 7 '19 at 20:16

Hint: $$\forall a>0,\; s\in(-1,1),\qquad J(a,s) = \int_{0}^{+\infty}\frac{x^{s}}{x^2+a^2}\,dx = \frac{\pi}{2\sin\frac{\pi s}{2}}\,a^{s-1}.$$ Consider the partial derivatives with respect to $a$ and $s$, then evaluate at $s=-\frac{1}{2}$.

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$$\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}=4 \int_{0}^\infty \frac{\log u \, du}{(u^4+a^2)^2}$$

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  • $\begingroup$ Thanks! I'll try to continue from there. $\endgroup$ – guest_user Jun 4 '16 at 11:36

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