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Compute $$I=\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$$

My attempt:

$$u=x^{1/3}\implies I = 3\int_0^{\infty}\frac {u}{u^6+1}du=\frac 32\int_0^\infty \frac {1}{u^3+1}du=\frac 32 \int_0^\infty \frac {1}{(x+1)(x^2+x+1)}dx\\\implies I=\frac 32 \int_0^{\infty} \frac 1{x+1}-\frac x{x^2+x+1}dx$$ And I'm stuck here, what can I do from here?

Larry
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C. Cristi
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  • Note that with the substitution $x\mapsto \sqrt{x}$, we see that $$\int_0^\infty \frac{1}{x^{1/3}(1+x^2)},dx=\frac12\int_0^\infty \frac{x^{-2/3}}{1+x},dx=\frac12B(1/3,2/3)=\frac12\Gamma(1/3)\Gamma(2/3)=\frac{\pi}{2\sin(\pi/3)}=\frac{\pi}{\sqrt 3}$$ – Mark Viola Jan 24 '19 at 21:18

5 Answers5

6

As an alternative approach:$$I=\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx\,\overset{\large x^{2/3}=u}=\,\frac32 \int_0^\infty \frac{1}{u^3+1}du$$ We will substitute $\displaystyle{u=\frac{1-t}{1+t}\Rightarrow du=-\frac{2}{(1+t)^2}dt}$. The reason behind it is that $(1-t)^3+(1+t)^3=2(3t^2+1)$, thus we get rid of the third powers. $$\Rightarrow I=\frac32 \int_{-1}^1 \frac{t+1}{3t^2+1}dt=\frac32 \cdot 2\int_0^1 \frac{dt}{3t^2+1}dt=3\cdot \frac{1}{\sqrt 3}\arctan(\sqrt 3 t)\bigg|_0^1=\frac{\pi}{\sqrt 3}$$

Zacky
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Hint. Taking from your last step, $$\int\frac x{x^2+x+1}dx=\frac{1}{2}\int\frac{D(x^2+x+1)}{x^2+x+1}dx-\frac{1}{2}\int\frac{1}{(x+1/2)^2+3/4}dx.$$ Can you take it from here?

Robert Z
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  • Okay so you added and $x+1$ and subtracted it? that's nice but after the calculation as the primitive I get :

    $$\ln (\sqrt{ \frac {(x+1^3)}{x^2+x+1} }) + \frac 1{\sqrt{3}} \arctan (\frac {(2x+1)}{\sqrt{3}})$$

     – C. Cristi Jan 24 '19 at 17:45
  • Yes. Now that you have the primitive evaluate it at 0 and the limit at $+\infty$ and take the difference. The primitive should be $3$ multiplied by $$\ln (\sqrt{ \frac {(x+1)^2}{x^2+x+1} }) + \frac 1{\sqrt{3}} \arctan (\frac {(2x+1)}{\sqrt{3}})$$ – Robert Z Jan 24 '19 at 17:53
  • What if there was actually $\int_0^\infty \frac 1{x^{2/3}(1+x^3)}dx$ What substitution would you use? – C. Cristi Jan 24 '19 at 18:04
  • I would try again $t=x^{1/3}$, but don't think that the primitive is "easy" here. Maybe it's better to change the approach and use complex analysis. – Robert Z Jan 24 '19 at 18:09
  • Already tried $x^{1/3}$ but it's clearly a lot harder, what kind of complex analysis? – C. Cristi Jan 24 '19 at 18:11
  • See this: https://math.stackexchange.com/questions/953364/how-to-integrate-int-0-infty-fracdx-sqrtxx21-using-the-resi – Robert Z Jan 24 '19 at 18:16
2

Also using the Beta function, but in a different way.

Recall that $$\mathrm{B}(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt$$ Using the substitution $w=\frac{1-t}t$, we see that $$\mathrm{B}(a,b)=\int_0^\infty\frac{w^{b-1}}{(1+w)^{a+b}}\mathrm dw$$ Then using the sub $w=u^2$, $$\mathrm{B}(a,b)=2\int_0^\infty\frac{u^{2b-1}}{(1+u^2)^{a+b}}\mathrm du$$ So setting $2b-1=-1/3$, and $a+b=1$, we see that $$\int_0^\infty \frac{\mathrm dx}{x^{1/3}(1+x^2)}=\frac12\mathrm{B}(1/3,2/3)$$

clathratus
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1

Quickest way would be to let $x\mapsto\tan x$ and use the beta function$$\begin{align*}\mathfrak{I} & =\int\limits_0^{\pi/2}\mathrm dx\,\tan^{-1/3}x\\ & =\int\limits_0^{\pi/2}\mathrm dx\,\sin^{-1/3}x\cos^{1/3}x\\ & =\frac 12\operatorname{B}\left(\frac 13,\frac 23\right)\end{align*}$$A direct application of Euler’s Reflection formula gives$$\int\limits_0^{\infty}\frac {\mathrm dx}{x^{1/3}(1+x^2)}\color{blue}{=\frac {\pi}{\sqrt3}}$$

Frank W
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1

Yet another approach.

\begin{align} \int_0^\infty \frac{dx}{\sqrt[3]{x} (1 + x^2)} &\overset{\large x \mapsto x^{3/2}}=\frac{3}{2} \int_0^\infty \frac{dx}{1 + x^3}\\ &= \frac{3}{2} \int_0^1 \frac{dx}{1 + x^3} + \frac{3}{2} \int_1^\infty \frac{dx}{1 + x^3}. \end{align} Enforcing a substitution of $x \mapsto 1/x$ in the second of the integrals leads to \begin{align} \int_0^\infty \frac{dx}{\sqrt[3]{x} (1 + x^2)} &= \frac{3}{2} \int_0^1 \frac{1 + x}{1 + x^3} \, dx\\ &= \frac{3}{2} \int_0^1 \frac{dx}{x^2 - x + 1}\\ &= \frac{3}{2} \int_0^1 \frac{dx}{\left (x - \frac{1}{2} \right )^2 + \frac{3}{4}}\\ &= \sqrt{3} \left [\tan^{-1} \left (\frac{2x - 1}{\sqrt{3}} \right ) \right ]_0^1\\ &= \frac{\pi}{\sqrt{3}}. \end{align}

omegadot
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