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Background: question 9.1.2.d in a newly published book Mathematics for Physical Science by Harris.

My question is what was the question supposed to be? Taking ln(x) for ln1 I'm doing something wrong or not guessing the right typo: $$\int_0^1(\ln(x)/x)^{1/3}\mathrm{dx}$$ $$x=e^{-u}$$ $$-\int_0^\infty (-u)^{1/3}*e^{-2u/3}(-\mathrm{du})$$ $$\int_0^\infty (-3/2u)^{1/3}*e^{-u}*(\frac{3}{2}\mathrm{du})$$

user5389726598465
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3 Answers3

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Notice that:

$$\int_0^\infty \frac{3}{2}\left(-3/2u\right)^{\frac{1}{3}}e^{-u}\mathrm{du} = \int_0^\infty \frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}u^{\frac{1}{3}}e^{-u}\mathrm{du} = \\ \frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\int_0^\infty u^{\frac{4}{3}-1}e^{-u}\mathrm{du} =\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{4}{3}\right). $$

Recall also that $$\Gamma(z) = (z-1)\Gamma(z-1).$$

Then:

$$\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{4}{3}\right) = \frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right) = \frac{1}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{1}{3}\right).$$

The results is confirmed also by Wolframalpha. Actually, you can't get rid of the multiplied term in front of the final expression.

the_candyman
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  • That's helpful, because it's close. I wonder if someone is clever enough to see by inspection how to get rid of the $(3/2)^{4/3}$ for instance like try $\ln(1/x)$ – user5389726598465 Apr 25 '17 at 09:32
  • Oh, I agree with your answer, I meant by inspection someone can see an alternative possibility for what the question was supposed to be. – user5389726598465 Apr 25 '17 at 09:44
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Why don't you write down the whole process to find out mistakes? The following is what I get:

$$I:=\int_0^1\left(\frac{\log x}x\right)^{1/3}dx\;:\;\;\text{substitution}\;\;e^{-u}:=x\implies-e^{-u}du=dx\implies$$

$$I=\int_\infty^0\left(\frac{(-u)}{e^{-u}}\right)^{1/3}(-e^{-u}du)=-\int_0^\infty u^{1/3}e^{-2u/3}du$$

and now a new substitution:

$$t=\frac{2u}3\implies du=\frac32dt\implies I=-\sqrt[3]\frac32\int_0^\infty t^{1/3}e^{-t}\frac32dt=-\sqrt[3]{\left(\frac32\right)^4}\,\Gamma\left(\frac43\right)$$

DonAntonio
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  • I've never taken a cube root of a negative sign... Maybe the error was supposed to be s.t. different... – user5389726598465 Apr 25 '17 at 09:40
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    @user135711 First of all, you always can, both within the real numbers or the complex ones: any odd root always exists for any real number. Second, this is not what I did above, but rather $$;u^{1/3}\to\left(\frac32t\right)^{1/3}=\sqrt[3]\frac32,t^{1/3};$$ – DonAntonio Apr 25 '17 at 09:42
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Starting with the gamma function $\int_0^\infty x^{1/3}e^{-x}$dx, and making the reverse substitution ln(1/x)=u (found by solving the original substitution $e^{-u}$) yields the intended question: $$\int_0^1(\ln(1/x))^{1/3}\mathrm{dx}$$ with solution $\Gamma(4/3)$

user5389726598465
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