Interesting integral 5

Question

Find an integral: $$ \int\limits_0^{+\infty} \frac{x^3 \ln^2 x}{1+x^7} d x. $$ Should I use residue theorem? May somebody suggest any ideas to find it?

Answer 1

We do this in three steps, using a pizza slice contour resting on the positive real axis and of radius $R$ with the angle being $2\pi/7.$ We start with

$$J_0 = \int_0^\infty \frac{x^3}{1+x^7} dx.$$

We get $J_0$ on the real axis. The circular arc is of the order $(2\pi/7) \times R \times \frac{R^3}{R^7}$ and hence makes no contribution. We get on the upper part parameterizing with $z=x \exp(2\pi i/7)$ a multiplier of $\exp(6\pi i/7) \exp(2\pi i/7).$ The only pole inside the slice is the one at $\rho = \exp(\pi i/7)$ and we get for the residue

$$\frac{\rho^3}{7\rho^6} = -\frac{\rho^4}{7}.$$

Hence by the Cauchy Residue Theorem we find that

$$J_0 (1-\exp(8\pi i/7)) = - 2\pi i \frac{\rho^4}{7}$$

so that

$$J_0 = - 2\pi i \frac{1}{7} \frac{\exp(4\pi i/7)}{1-\exp(8\pi i/7)} = - 2\pi i \frac{1}{7} \frac{1}{\exp(-4\pi i/7)-\exp(4\pi i/7)} \\ = \frac{1}{7} \frac{\pi}{\sin(4\pi/7)}.$$

Next we continue with

$$J_1 = \int_0^\infty \frac{x^3 \log x}{1+x^7} dx$$

where we indent the pizza slice around the origin. The large circular arc makes no contribution in the limit, same as before. The small arc is dominated by $2\pi \epsilon/7 \times \epsilon^3 |\log\epsilon + 2\pi i|$ which vanishes as $\epsilon\rightarrow 0.$

We get from the upper part of the slice

$$- J_0 \exp(8\pi i/7) \times 2\pi i/7 - J_1 \exp(8\pi i/7).$$

The residue now becomes

$$-\frac{\exp(4\pi i/7)}{7} (\pi i/7)$$

We thus have

$$J_1 (1-\exp(8\pi i/7)) = J_0 \exp(8\pi i/7) \times 2\pi i/7 - 2\pi i \frac{\exp(4\pi i/7)}{7} (\pi i/7).$$

This yields

$$J_1 = J_0 \frac{\exp(8\pi i/7)}{1-\exp(8\pi i/7)} \times 2\pi i/7 + J_0 \frac{\pi i}{7} \\ = J_0 \frac{\pi i}{7} \frac{1+\exp(8\pi i/7)}{1-\exp(8\pi i/7)} = - J_0 \frac{\pi i}{7} \frac{2\cos(4\pi/7)}{2i\sin(4\pi/7)} = - J_0 \frac{\pi}{7} \cot(4\pi /7) \\ = - \frac{\pi^2}{7^2} \frac{\cot(4\pi /7)}{\sin(4\pi/7)}.$$

To conclude we treat the target integral which is

$$J_2 = \int_0^\infty \frac{x^3 \log^2 x}{1+x^7} dx.$$

We use $(\log x + 2\pi i/7)^2 = \log^2 x + 4\pi i/7 \log x - 4\pi^2/7^2$ and obtain on the upper part (same contour, same estimates on the arcs)

$$J_0 \exp(8\pi i/7) \frac{4\pi^2}{7^2} - J_1 \exp(8\pi i/7) \frac{4\pi i}{7} - J_2 \exp(8\pi i/7)$$

so that

$$J_2 = J_0 \frac{\exp(8\pi i/7)}{1-\exp(8\pi i/7)} \left(-\frac{4\pi^2}{7^2} - \frac{4\pi^2 i}{7^2}\cot(4\pi/7)\right) - J_0 \frac{\pi^2}{7^2} \\ = -\frac{4\pi^2}{7^2} J_0 \frac{\exp(8\pi i/7)}{1-\exp(8\pi i/7)} (1+i\cot(4\pi/7)) - J_0 \frac{\pi^2}{7^2}.$$

To simplify this we momentarily put $z = \exp(4\pi i/7)$ and get for the central term

$$\frac{z^2}{1-z^2} \left(1+i\frac{(z+1/z)/2}{(z-1/z)/2/i}\right) = \frac{z^2}{1-z^2} \left(1- \frac{z+1/z}{z-1/z}\right) \\ = -\frac{z^2}{1-z^2} \frac{2/z}{z-1/z} = -\frac{z^2}{1-z^2} \frac{2}{z^2-1} = \frac{2z^2}{(z^2-1)^2} = \frac{2}{(z-1/z)^2}.$$

Putting it all together we obtain

$$-\frac{\pi^3}{7^3} \frac{1}{\sin(4\pi/7)} \left(4 \times -\frac{1}{2} \frac{1}{\sin^2(4\pi/7)} + 1\right) \\ = \frac{\pi^3}{7^3} \frac{1}{\sin(4\pi/7)} \left(\frac{2}{\sin^2(4\pi/7)} - 1\right) = \frac{\pi^3}{7^3} \frac{2-\sin^2(4\pi/7)}{\sin^3(4\pi/7)}.$$

We conclude that the integral is

$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi^3}{7^3} \frac{1+\cos^2(4\pi/7)}{\sin^3(4\pi/7)}.}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{equation} \mbox{Note that}\quad \int_{0}^{\infty}{x^{3}\ln^{2}\pars{x} \over 1 + x^{7}}\,\dd x = \left.\partiald[2]{}{\mu}\int_{0}^{\infty}{x^{\mu} \over 1 + x^{7}}\,\dd x \,\right\vert_{\ \mu\ =\ 3} \label{1}\tag{1} \end{equation}

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With $\ds{x \equiv \pars{{1 \over t} - 1}^{1/7} \implies t = {1 \over 1 + x^{7}}}$: \begin{align} \int_{0}^{\infty}{x^{\mu} \over 1 + x^{7}}\,\dd x & = \int_{1}^{0}t\pars{{1 \over t} - 1}^{\mu/7} \bracks{{1 \over 7}\pars{{1 \over t} - 1}^{-6/7}\pars{-\,{1 \over t^{2}}}}\dd t \\[6mm] & = {1 \over 7}\int_{0}^{1}t^{-\pars{\mu + 1}/7}\pars{1 - y}^{\pars{\mu - 6}/7} \,\dd t = {1 \over 7}\,{\Gamma\pars{\bracks{6 - \mu}/7}\Gamma\pars{\bracks{\mu + 1}/7} \over \Gamma\pars{1}} \\[5mm] & = {1 \over 7}\,{\pi \over \sin\pars{\pi\bracks{\mu + 1}/7}} = {1 \over 7}\,\pi\csc\pars{{\pi \over 7}\,\bracks{\mu + 1}} \end{align}

After a 'laborious' procedure; I'll find, by replacing in \eqref{1},

$$\bbx{\ds{% \int_{0}^{\infty}{x^{3}\ln^{2}\pars{x} \over 1 + x^{7}}\,\dd x = {\pi^{3} \over 686}\bracks{3 - \cos\pars{\pi \over 7}} \sec^{3}\pars{\pi \over 14}}} \approx 0.1024 $$