5

Came across

$$I=\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$$

and broke the integrand as

$$\frac{1-x(2-\sqrt x)}{1-x^3}=\frac{1-x}{1-x^3}- \frac{x(1-\sqrt x)}{1-x^3} $$ The first term simplifies and the second term transforms with $\sqrt x\to x$. Then, the integral becomes

$$I=\int_0^\infty \frac{1}{1+x+x^2}dx - 2\int_0^\infty \frac{x^3}{1+x+x^2+x^3+x^4+x^5}dx $$ The first integration is straightforward $\frac{2\pi}{3\sqrt3}$ and the second is carried out via partial fractional decomposition. After some tedious and lengthy evaluation, it surprisingly yields the same value $\frac{2\pi}{3\sqrt3}$.

Given its vanishing value, there may exist a shorter and more direct derivation, without evaluation of any explicit intermediate values.

Integreek
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Quanto
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2 Answers2

3

$$\require{cancel}\begin{align}\int_0^\infty\frac{x^\frac32-2x+1}{1-x^3}\mathrm dx&=\int_0^\infty\frac{-x+\sqrt x+1}{(\sqrt x+1)(x^2+x+1)}\mathrm dx\\&\overset{t=\sqrt x}{=}2\int_0^\infty\frac{t(-t^2+t+1)}{(t+1)(t^4+t^2+1)}\mathrm dt\\&=-2\int_0^\infty\frac{t\,\mathrm dt}{(t+1)(t^2+t+1)}+4\int_0^\infty\frac{t\,\mathrm dt}{(t+1)(t^4+t^2+1)} \\&\overset{t\to\frac1t}{=}-2\int_0^\infty\frac{t\,\mathrm dt}{(t+1)(t^2+t+1)}+2\int_0^\infty\frac{t\,\mathrm dt}{t^4+t^2+1}\\&=-2\int_0^\infty\frac{t^3-2t^2}{(t+1)(t^4+t^2+1)}\mathrm dt\\&\overset{t\to\frac1t}{=}-\int_0^\infty\frac{t^2-3t+1}{t^4+t^2+1}\mathrm dt\\&=\cancel{\int_0^\infty\frac{\mathrm d(t^2)}{t^4+t^2+1}}-\cancel{\int_0^\infty\frac{\mathrm dt}{t^2+t+1}}\\&=0\end{align}$$

Integreek
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1

I'm not sure if this is something that is of value to you since it uses your approach, but once you transform $\sqrt{x} \to x$ for the second integral you can use PFD to cancel out the first integral and be left with two simple integrals, instead of expressing it as you did: $$-\int_0^{\infty} \frac{2x^3(1-x)}{1-x^6} \; \mathrm{d}x=-\int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\frac{1}{3}\int_0^{\infty} \frac{2x-1}{x^2-x+1} \; \mathrm{d}x+\frac{2}{3} \int_0^{\infty} \frac{\mathrm{d}x}{x+1} $$ And so you have, \begin{align*} I&= \int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\frac{1}{3}\int_0^{\infty} \frac{2x-1}{x^2-x+1} \; \mathrm{d}x+\frac{2}{3} \int_0^{\infty} \frac{\mathrm{d}x}{x+1} \\ &= \bigg [-\frac{1}{3} \ln(x^2-x+1)+\frac{2}{3}\ln(x+1) \bigg]\bigg \rvert_0^{\infty} \\ &=0 \end{align*}

Ty.
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