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I found the method of partial fractions very laborious to solve this definite integral : $$\int_0^\infty \frac{\sqrt[3]{x}}{1 + x^2}\,dx$$

Is there a simpler way to do this ?

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    $\begingroup$ You could use a keyhole contour, (see en.wikipedia.org/wiki/…) and choose $[0,\infty)$ as the branch cut for Log. However this may or may not be "simpler" depending on your definition, and how comfortable you are with complex analytic techniques. $\endgroup$ Apr 21, 2011 at 19:13
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    $\begingroup$ Thanks. That was helpful. I haven't studied contour integration yet, but now I know I should. $\endgroup$
    – Balaji Rao
    Apr 21, 2011 at 19:26
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    $\begingroup$ Sorry to be picky, but you don't solve integrals, you evaluate them. $\endgroup$ Jun 9, 2012 at 2:18
  • $\begingroup$ Related to: math.stackexchange.com/questions/373164/… $\endgroup$ Apr 26, 2013 at 5:57

8 Answers 8

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Perhaps this is simpler.

Make the substitution $\displaystyle x^{2/3} = t$. Giving us

$\displaystyle \frac{2 x^{1/3}}{3 x^{2/3}} dx = dt$, i.e $\displaystyle x^{1/3} dx = \frac{3}{2} t dt$

This gives us that the integral is

$$I = \frac{3}{2} \int_{0}^{\infty} \frac{t}{1 + t^3} \ \text{d}t$$

Now make the substitution $t = \frac{1}{z}$ to get

$$I = \frac{3}{2} \int_{0}^{\infty} \frac{1}{1 + t^3} \ \text{d}t$$

Add them up, cancel the $\displaystyle 1+t$, write the denominator ($\displaystyle t^2 - t + 1$) as $\displaystyle (t+a)^2 + b^2$ and get the answer.

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  • $\begingroup$ Nice solution, +1. Though what you did is much cleaner, it is interesting to note that with some work you can actually write down the exact anti derivative for $\frac{1}{1+x^3}.$ (Of course it uses partial fractions, which the OP didn't want, and has all a few logs and artans and is a bit messy) $\endgroup$ Apr 21, 2011 at 19:57
  • $\begingroup$ Thank you. That was an excellent solution! $\endgroup$
    – Balaji Rao
    Apr 22, 2011 at 4:48
  • $\begingroup$ Very nice solution, +1. $\endgroup$ May 6, 2011 at 10:05
  • $\begingroup$ Very clever, congratulations! $\endgroup$ Apr 8, 2012 at 19:20
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By using techniques of complex analysis ($\text{Residue Theory}$) one can actually show that $$\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \qquad 0 < a <b$$

You can obtain the value of your $\text{Integral}$ by putting $a=\frac{4}{3}$ and $b=2$.

Set $$I = \int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx}$$ and integrate $$f(z) = \frac{z^{a-1}}{1+z^{b}} = \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}}$$

Simple pole at $z_{1} = e^{\pi{i}/b}$ and hence $$\text{Res} \Biggl[\frac{z^{a-1}}{1+z^{b}}, e^{\pi{i}/b}\Biggr] = \frac{z^{a-1}}{bz^{b-1}}\Biggl|_{z =e^{\pi i / b}} = -\frac{1}{b}e^{\pi i a/b}$$

Integrate along $\gamma_{1}$, and let $R \to \infty$ and let $ \epsilon \to 0^{+}$. This gives, \begin{align*} \int\limits_{\gamma_{1}} f(z) \ dz & = \int\limits_{\gamma_{1}} \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}} \ dz \\ &= \int\limits_{\epsilon}^{R} \frac{x^{a-1}}{1+x^{b}} \to \int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ dx =I \end{align*}

Integrate along $\gamma_{2}$, and let $R \to \infty$. This gives $0 < a < b$ and $$\Biggl|\int\limits_{\gamma_{2}} f(z) dz \Biggr| \leq \frac{R^{a-1}}{R^{b}-1} \cdot \frac{2\pi R}{b} \sim \frac{2 \pi}{b R^{b-a}} \to 0$$

Integrate along $\gamma_{3}$ and let $R \to \infty$ and $\epsilon \to 0^{+}$. This gives \begin{align*} \int\limits_{\gamma_{3}} f(z) \ dz &= \int\limits_{\gamma_{3}} \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}} = \Biggl[\begin{array}{c} z=x e^{2\pi i/b} \\ dz=e^{2\pi i/b} \ dx \end{array}\Biggr] \\ &= \int\limits_{R}^{\epsilon} \frac{x^{a-1}e^{2\pi i(a-1)/b}}{1+x^{b}} \cdot e^{2\pi i b} \ dx \to \int\limits_{\infty}^{0} \frac{x^{a-1}e^{2\pi i(a-1)/b}}{1+x^{b}} \cdot e^{2\pi i b} \ dx \\ &= -e^{2\pi ia/b}I \end{align*}

Integrate along $\gamma_{4}$ and let $\epsilon \to 0^{+}$. This gives $0 < a <b$, $$\Biggl|\int\limits_{\gamma_{4}} f(z) \ dz \Biggr| \leq \frac{\epsilon^{a-1}}{1-\epsilon^{b}} \cdot \frac{2\pi\epsilon}{b} \sim \frac{2\pi\epsilon}{b} \to 0$$

Using the $\text{Residue Theorem}$ and letting $R \to \infty$ and $\epsilon \to 0^{+}$, we obtain that $$ I + 0 - e^{2\pi a/b}I + 0 = 2\pi i \cdot \Bigl(-\frac{1}{b} e^{\pi ia/b}\Bigr)$$ This yields, $$(e^{-\pi i a/b} - e^{\pi i a./b})I= -\frac{2\pi i}{b}$$ and hence solving for $I$, we have $$I= \frac{2\pi i}{b \cdot (e^{\pi ia/b} - e^{-\pi i a/b})}=\frac{\pi}{b \sin(\pi a/b)}$$

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  • $\begingroup$ Heh, I actually started out on this problem in trying to prove the reflection formulas for the Gauss Pi function and the Legendre Gamma function: $\Pi(z)\Pi(-z) = \frac{\pi z}{\sin \pi z}$ $\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin \pi z}$ So using the Beta and Gamma functions would be circular for what I'm trying to do. But interesting that you would come to what amounts to the reverse. I had thought there'd be a method using the residue theorem but I wasn't sure what contour to use. Chandrasekhar, I'm still not clear on your choice of contours...what are $\gamma_1$ and $\gamma_2$? $\endgroup$
    – Eric
    Feb 29, 2012 at 20:39
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    $\begingroup$ Standard reverse keyhole contour that bypass the origin and wrap up the positive part of the real line twice. $\endgroup$ Sep 1, 2012 at 7:58
  • $\begingroup$ I'm not sure, @user32240, that user9413 meant that contour, as there doesn't seem to be any reason to to bypass the origin...It's a rather unusual and odd omission not to specify clearly what the contour is in such an answer. $\endgroup$
    – DonAntonio
    Sep 2, 2012 at 2:22
  • $\begingroup$ Well, after seeing the integrals $\,\int_\epsilon^R\,$ and etc. I think you're right...why was the origin bypassed?? I don't understand $\endgroup$
    – DonAntonio
    Sep 2, 2012 at 2:28
  • $\begingroup$ @DonAntonio only stumbling across this answer now, which is long after your comments; is it because $\arg(z)$ isn't well defined at the origin that you bypass the origin? $\endgroup$
    – snulty
    Aug 12, 2015 at 3:03
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Here is a different way I am quite fond of:

Call our integral I, that is set $$I=\int_0^\infty \frac{\sqrt[3]{x}}{1+x^2}\,dx$$ Let $u=1+x^{2}$ so that $du=2x \, dx$. Since $$\sqrt[3]{x} \, dx=\frac{1}{2}\frac{2x \, dx}{\sqrt[3]{x^{2}}}=\frac{1}{2}\frac{u}{\left({u-1}\right)^{\frac{1}{3}}}\,du$$ we have that$$I=\frac{1}{2}\int_{1}^{\infty}\frac{1}{u\left(u-1\right)^{\frac{1}{3}}}\,du.$$ Let $u=\frac{1}{v}$ so that this becomes $$\frac{1}{2}\int_{0}^{1}\frac{1}{\frac{1}{v}\left(\frac{1}{v}-1\right)^{\frac{1}{3}}}\frac{1}{v^{2}}\,dv=\frac{1}{2}\int_{0}^{1}v^{-\frac{2}{3}}\left(1-v\right)^{-\frac{1}{3}}\,dv=\frac{1}{2}\text{B}\left(\frac{1}{3},\frac{2}{3}\right)$$ where $\text{B}(x,y)$ is the beta function. Since $$\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ we have that $$I=\frac{1}{2}\frac{\Gamma(\frac{1}{3})\Gamma(\frac{2}{3})}{\Gamma(1)}.$$ Since $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}$ it follows that $$I=\frac{\pi}{2\sin\pi/3}=\frac{\pi}{\sqrt{3}}.$$ Hope that helps,

Also it is worth mentioning that numerically, Wolfram Alpha agrees with this answer.

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    $\begingroup$ Nice! Two quibbles: add $\mathrm{d}u$ in your first change of variables; and the established notation for the Beta function is B (standing for a capital beta). $\endgroup$
    – Did
    Apr 21, 2011 at 20:08
  • $\begingroup$ Maybe you want to include after stating the Beta function, that it contains the gamma function which is in the fractions so people can recognize the properties that the gamma function holds. $\endgroup$
    – night owl
    May 6, 2011 at 9:54
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    $\begingroup$ @owl: The "Since $B(x,y)=\dots$" looks pretty explicit to me. $\endgroup$ May 6, 2011 at 10:25
  • $\begingroup$ I have used both this method and residues for this kind of problem. I like them both! (+1) $\endgroup$
    – robjohn
    Aug 11, 2012 at 13:41
  • $\begingroup$ It looks like we use a similar approach. +1. $\endgroup$
    – Tunk-Fey
    May 31, 2014 at 14:16
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Using the same technique as in my previous answer we can generalize, and find the Mellin Transform:

Consider $$I(\alpha,\beta)=\int_{0}^{\infty}\frac{u^{\alpha-1}}{1+u^{\beta}}du=\mathcal{M}\left(\frac{1}{1+u^{\beta}}\right)(\alpha)$$ Let $x=1+u^{\beta}$ so that $u=(x-1)^{\frac{1}{\beta}}$. Then we have $$I(\alpha,\beta)=\frac{1}{\beta}\int_{1}^{\infty}\frac{(x-1)^{\frac{\alpha-1}{\beta}}}{x}(x-1)^{\frac{1}{\beta}-1}dx.$$ Setting $x=\frac{1}{v}$ we obtain $$I(\alpha,\beta)=\frac{1}{\beta}\int_{0}^{1}v^{-\frac{\alpha}{\beta}}(1-v)^{\frac{\alpha}{\beta}-1}dv=\frac{1}{\beta}\text{B}\left(-\frac{\alpha}{\beta}+1,\ \frac{\alpha}{\beta}\right).$$

Using the properties of the Beta and Gamma functions, this equals $$\frac{1}{\beta}\frac{\Gamma\left(1-\frac{\alpha}{\beta}\right)\Gamma\left(\frac{\alpha}{\beta}\right)}{\Gamma(1)}=\frac{\pi}{\beta\sin\left(\frac{\pi\alpha}{\beta}\right)}.$$

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Let's generalize the problem. We will evaluate $$ \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx. $$ Let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\large\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then \begin{align} \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\large\frac{a-1}b}\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\\&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy, \end{align} where the last integral in RHS is Beta function. $$ \text{B}(x,y)=\int_0^1t^{\ \large x-1}\ (1-t)^{\ \large y-1}\ dt=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}. $$ Hence \begin{align} \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy\\&=\frac1b\cdot\Gamma\left(1-\frac{a}{b}\right)\cdot\Gamma\left(\frac{a}{b}\right)\\&=\large{\color{blue}{\frac{\pi}{b\sin\left(\frac{a\pi}{b}\right)}}}. \end{align} The last part uses Euler's reflection formula for Gamma function provided $0<a<b$. Thus $$ \large\int_0^\infty\dfrac{\sqrt[3]{x}}{1+x^2}\ dx=\int_0^\infty\dfrac{x^{\large\frac43-1}}{1+x^2}\ dx=\frac{\pi}{2\sin\left(\frac{2\pi}{3}\right)}=\color{blue}{\frac\pi3\sqrt{3}}. $$

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We can use the following results $$\sum_{n=-\infty}^\infty(-1)^n\frac{1}{bn+a}=\frac{\pi}{b\sin\frac{a\pi}{b}}, \frac{1}{1-x}=\sum_{n=0}^\infty x^n $$ to evaluate the generalization. In fact \begin{eqnarray} \int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=&\int_0^1\frac{x^{a-1}}{1+x^b}dx+\int_0^1\frac{x^{-a-1}}{1+x^b}dx\\ &=&\sum_{n=0}^\infty(-1)^n\int_0^1(x^{bn+a-1}+x^{bn-a-1})dx\\ &=&\sum_{n=0}^\infty(-1)^n(\frac{1}{bn+a}+\frac{1}{bn-a})\\ &=&\sum_{n=-\infty}^\infty(-1)^n\frac{1}{bn+a}\\ &=&\frac{\pi}{b\sin\frac{a\pi}{b}}. \end{eqnarray}

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We can convert the integral into a more familiar $\sin(x)/x$ type form using $$ I=\int_0^\infty f(x) g(x)\; dx = \int_0^\infty \mathcal{L}[f(x)](s)\mathcal{L}^{-1}[g(x)](s)\;ds $$ $$ \mathcal{L}[x^{1/3}](s)=\frac{\Gamma(4/3)}{s^{4/3}} $$ $$ \mathcal{L}^{-1}\left[\frac{1}{1+x^2}\right](s) =\sin(s) $$ So $$ \int_0^\infty \frac{x^{1/3}}{1+x^2}\;dx =\Gamma\left(\frac{4}{3}\right)\int_0^\infty \frac{\sin(x)}{x^{4/3}}\;dx = \frac{3}{2}\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right) = \frac{\pi}{\sqrt{3}} $$

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Another way

Let

$$ I= \int_0^\infty \frac {x^{1/3}\, \mathrm{d}x}{x^2 + 1}. $$

By the Schwinger parametrization we have

$$ I=\int_0^\infty \mathrm{d}t\, \exp(-t)\int_0^\infty \mathrm{d}x\, x^{1/3}\, \exp\left(-tx^2\right).$$

The last integral can be calculated along the lines of this answer. Using this result, one gets

$$I=\int_0^\infty \mathrm{d}t\,\frac{ \Gamma \left(\frac{2}{3}\right)\exp(-t)}{2 t^{2/3}}=\frac{1}{2}\Gamma \left(\frac{2}{3}\right)\Gamma \left(\frac{1}{3}\right)=\frac{\pi}{\sqrt{3}},$$

since

$$ \Gamma\left(\nu,x\right)\equiv\int_x^\infty \mathrm{d}t\, t^{\nu-1}\exp(-t).$$

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