evaluate the improper integral of $\frac{x^{-\frac{1}{3}}}{1+x}$

Question

Evaluate

$$\int_0^\infty \frac{x^{-\frac{1}{3}}}{x+1}dx$$

If I am using complex analysis is $f(x)=x^{-\frac{1}{3}}$ my function? And there are no zeroes or poles on $(0,\infty)$ since $x=-1 \not \in (0,\infty)$ is the only point of interest. Any hints greatly appreciated. I can write the integrand as

$$\frac{1}{x^{4/3}+x^{1/3}}$$ how do I integrate this??

@J.G. the numerator was $x^{-1/3}$ not positive. does this change it? — homosapien, Feb 01 '22 at 21:13
@user how do I go about proving this? so I use Cauchy's formula ? Or brut force? — homosapien, Feb 01 '22 at 21:14
@user How do I integrate it from this step $\int_0^\infty \frac{1}{x^{4/3}+x^{1/3}}dx$ — homosapien, Feb 01 '22 at 21:19
Probably you know that you should apply complex analysis. Otherwise you would not mention it in the question, would you? — user, Feb 01 '22 at 21:21
Does this answer your question? What is $\int_0^\infty\frac{x^{z-1}}{x+1} dx$? (specifically this answer) — metamorphy, Feb 02 '22 at 07:23
@metamorphy that one's slightly different, I got this one answered using substitutions!! thanks though! :) cheers!!! — homosapien, Feb 02 '22 at 17:51
But this answer does not use complex analysis which you mentioned. — user, Feb 03 '22 at 07:06

score 1 · Accepted Answer · edited Feb 02 '22 at 17:47

1

$$\int_0^\infty \frac{x^{-\frac{1}{3}}}{x+1}dx\,\,\,\,\stackrel{x=u^3}{=}\,\,\,\,\int_0^\infty \frac{3u}{u^3+1}du\,\,\,\,\stackrel{v=\frac 1 u}{=}\,\,\,\,\int_0^\infty \frac{3}{v^3+1}dv$$ Thus, summing the last 2 expressions, $$\begin{split} 2\int_0^\infty \frac{x^{-\frac{1}{3}}}{x+1}dx&=3\int_0^\infty \frac{v+1}{v^3+1}dv\\ &=\int_0^\infty \frac{3}{1-v+v^2}dv\\ &=\int_0^\infty \frac{4}{1+\frac 4 3\left(v-\frac 1 2\right)^2}dv\\ &= \left[2\sqrt{3}\arctan \left(\frac{2v-1}{\sqrt 3}\right)\right]_0^{+\infty}\\ &=2\sqrt{3}\left(\frac\pi 2+\frac \pi 6\right)\\ &=\frac{4\pi}{\sqrt 3} \end{split}$$

Thus the integral $\int_0^\infty \frac{x^{-\frac{1}{3}}}{x+1}dx = \frac{1}{2} \cdot \frac{4 \pi}{\sqrt{3}}=\frac{2 \pi}{\sqrt{3}}$.

edited Feb 02 '22 at 17:47

homosapien

4,157

answered Feb 02 '22 at 07:13

Stefan Lafon

12,256
11
29

why are you summing the last two expressions? and where did the $2$ come from ? – homosapien Feb 02 '22 at 17:22
also, a bit confused on the $v$ substitution. – homosapien Feb 02 '22 at 17:26
So we found two alternate expressions for the same integral (that's what the first line is about). The next step is to say that twice the original integral is the sum of those two expressions. – Stefan Lafon Feb 02 '22 at 17:41
so you rewrote the integral as two distinct integrals and then shouldn't the answer be half of what you have then? So you had it as two integrals, so you wrote twice the integral as the sum of the two "version" of the original integral? – homosapien Feb 02 '22 at 17:43
Also, could you elaborate on how you did the $v$ substitution, I don't see how the $u$ vanishes on top but the denominator stays the same – homosapien Feb 02 '22 at 17:44
1

Yes, twice the original integral is equal to $\frac {4\pi}{\sqrt 3}$, so the original integral itself is equal to half of that, i.e. $\frac {2\pi}{\sqrt 3}$. – Stefan Lafon Feb 02 '22 at 17:47
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For the $v$ substitution, I recommend you write it down on paper: it's equal to $\int_0^\infty \frac{3v^{-1}}{v^{-3}+1}\frac{dv}{v^2}=\int_0^\infty \frac{3}{v^3+1}dv$. – Stefan Lafon Feb 02 '22 at 17:49
I tried writing t down, got messy, ahh I see thanks !!! – homosapien Feb 02 '22 at 17:50
I keep getting $du=-\frac{1}{v^2} dv$ where did your negative go? sorry for bothering. – homosapien Feb 02 '22 at 18:08
where did the negative go? – homosapien Feb 02 '22 at 18:43
Let us continue this discussion in chat. – Stefan Lafon Feb 03 '22 at 00:42

score 0 · Answer 2 · answered Feb 01 '22 at 21:25

0

This should be a duplicate, but I can't find where this, or a similar more general integral, is evaluated by the desired methods, so let me give a hint that's unfortunately too long for a comment. For small $\epsilon>0$ and large $R>1$, let $\gamma$ be the contour that moves in a line segment from $i\epsilon$ to $R+i\epsilon$, in a near circle to $R-i\epsilon$, in a line segment from $R-i\epsilon$ to $-i\epsilon$, and in a semicircle from $-i\epsilon$ to $i\epsilon$. By the residue theorem,$$\oint_\gamma\frac{\exp(-\frac13\ln z)dz}{1+z}=2\pi i\left.\exp\left(-\frac13\ln z\right)\right|_{z=-1}$$(I leave it to you to work out which branch of the complex logarithm should be used here). Now show the contribution of the semicircle (near-circle) vanishes as $\epsilon\to0^+$ ($R\to\infty$). The rest is$$\left(1-\exp\left(-\frac{2\pi i}{3}\right)\right)\int_0^\infty\frac{\exp(-\frac13\ln x)}{1+x}dx,$$because rotating through $2\pi$ radians to get to the second line segment, thereby multiplying $z$ by $e^{2\pi i}=1$ (so the integrand's denominator is unchanged), rescales the integrand. (The $-$ sign is beause we run leftward on the second line segment.)

If this contour shape is unfamiliar, note this is similar to $\int_{-\infty}^\infty f(x)dx=2\int_0^\infty f(x)dx$ for even $f$. (The much more famous infinite semicircular contour can be thought of in similar terms, if you go from $0$ to $Re^{i0}$ to $Re^{i\pi}$ to $0$.)

answered Feb 01 '22 at 21:25

J.G.

115,835

are we using the principle branch of the logarithm? as $z=-1$ we cannot have in the log. By principle brand I mean putting out the negative real axis, so $\mathbb{C} \setminus \mathbb{R}^{\leq 0}$ – homosapien Feb 01 '22 at 21:27
@HossienSahebjame If we should take $\ln(-1)=\pi i$, or $\ln(-1)=3\pi i$, or whatever, you can work out a value for your original integral. Can you see which branch makes the answer real? Can you see another reason to accept it? (It's related to the rotating-round-the-circle argument later in my answer.) – J.G. Feb 01 '22 at 21:30
So its the branch excluding the positive reals ? – homosapien Feb 01 '22 at 21:31
@HossienSahebjame I'm not sure what you mean by that. A branch choice in this case means deciding which integer $n$ in $\ln(-1)=\pi i+2n\pi i$ suits here. – J.G. Feb 01 '22 at 21:32
so an $n$ that lands us on the real axis then? – homosapien Feb 01 '22 at 21:33
is it $n=0$? I just competed it out – homosapien Feb 01 '22 at 21:35

score 0 · Answer 3 · answered Feb 01 '22 at 22:02

Well, we are trying to solve the following improper integral:

$$\mathcal{I}_\text{n}\left(\alpha\right):=\int_0^\infty\frac{x^\text{n}}{x+\alpha}\space\text{d}x\tag1$$

Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:

$$\mathcal{I}_\text{n}\left(\alpha\right)=\int_0^\infty\mathscr{L}_x\left[x^\text{n}\right]_{\left(\sigma\right)}\cdot\mathscr{L}_x^{-1}\left[\frac{1}{x+\alpha}\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag2$$

Using the table of selected Laplace transforms, we can see:

$$\mathcal{I}_\text{n}\left(\alpha\right)=\Gamma\left(1+\text{n}\right)\int_0^\infty\frac{\exp\left(-\alpha\sigma\right)}{\sigma^{1+\text{n}}}\space\text{d}\sigma\tag3$$

Using the sum definition of the $\exp$ function, we can write:

$$\mathcal{I}_\text{n}\left(\alpha\right)=\Gamma\left(1+\text{n}\right)\lim_{\text{m}\space\to\space\infty}\sum_{\text{k}\space\ge\space0}\frac{\left(-1\right)^\text{k}\alpha^\text{k}}{\text{k}!}\int_0^\text{m}\sigma^{\text{k}-\text{n}-1}\space\text{d}\sigma\tag4$$

So, we get:

$$\mathcal{I}_\text{n}\left(\alpha\right)=\Gamma\left(1+\text{n}\right)\lim_{\text{m}\space\to\space\infty}\sum_{\text{k}\space\ge\space0}\frac{\left(-1\right)^\text{k}\alpha^\text{k}}{\text{k}!}\cdot\frac{\text{m}^{\text{k}-\text{n}}}{\text{k}-\text{n}}\tag5$$

Now, when $\text{n}=-\frac{1}{3}$ and $\alpha=1$ we get:

$$\mathcal{I}_{-\frac{1}{3}}\left(1\right)=\Gamma\left(\frac{2}{3}\right)\lim_{\text{m}\space\to\space\infty}\sum_{\text{k}\space\ge\space0}\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{\text{m}^{\text{k}+\frac{1}{3}}}{\text{k}+\frac{1}{3}}=\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)=\frac{2\pi}{\sqrt{3}}\tag6$$

evaluate the improper integral of $\frac{x^{-\frac{1}{3}}}{1+x}$

3 Answers3