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$$\int_0^\infty x^{1/2}e^{-x^3}\,dx=I$$ $$J=\int_0^1x^{-2/3}(1-x)^{-1/3}\,dx$$ $$J/I=?$$ I am confused how to find the value of $I$ and $J$. I tried to find them separately but couldn't.

I think by using some substitution in $J$ I shall get some scalar multiple of $I$, hence obtaining the required value.

I am stuck. Guide me!

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    Are you familiar with the Gamma and Beta functions? – 2017-01-13
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    No I M not. Can't it be solved without using that? – 2017-01-13
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    Technically, you can solve it without those functions, but why would you want to? – 2017-01-13
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    I don't understand these type of exercises. The two integrals have nothing to do with each other, it seems. Anyways, you have a way of calculating the first one in one answer, the second one looks easier. – 2017-01-13
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    @mickep: both integrals are related with the values of the $\Gamma$ function at $\frac{1}{3},\frac{1}{2},\frac{2}{3}$, and I do not agree about the second integral being "easier" than the first one! – 2017-01-13
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    Well, the second one has an elementary primitive... – 2017-01-13
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    @JackD'Aurizio The second integral can be transformed into an integral of a rational function, making it elementary. That's one definition of "easier". :) – 2017-01-13
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    @JackD'Aurizio See my hint below... – 2017-01-13

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Here's a hint for solving the second integral without beta function machinery:

$$\begin{align} J &=\int_{0}^{1}x^{-2/3}\left(1-x\right)^{-1/3}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt[3]{x^{2}\left(1-x\right)}}\\ &=\int_{0}^{\infty}\frac{\mathrm{d}u}{u^{2/3}\left(1+u\right)};~~~\small{\left[x=\frac{u}{1+u}\right]}\\ &=3\int_{0}^{\infty}\frac{\mathrm{d}t}{1+t^{3}};~~~\small{\left[\sqrt[3]{u}=t\right]}.\\ \end{align}$$

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    (+1) This is more or less what I meant. It is easier, since one can give it to a first-year calculus student. – 2017-01-13
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Hint. By the change of variable $t=x^3$, $dx=\frac13t^{-2/3}dt$, one gets $$ \int_0^\infty x^{1/2}e^{-x^3}dx=\frac13\int_0^\infty t^{1/2-1}e^{-t}dt=\frac23\int_0^\infty e^{-u^2}du $$ then one may conclude with the Euler gamma function or with the gaussian integral.

Concerning $J$ use the Euler beta function result.

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    please check my question http://math.stackexchange.com/questions/2099282/counting-the-number-of-partition-using-bijection?noredirect=1#comment4316053_2099282 – 2017-01-15