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If $\displaystyle I=\int^{\infty}_{0}e^{-2x}\cdot x^6dx$ and $\displaystyle J=\int^{2}_{0}x(8-x^3)^{\frac{1}{3}}dx$

Then product of $I$ and $J$ equals

Try: For $\displaystyle I = \int^{\infty}_{0}e^{-2x}\cdot x^6dx = \frac{1}{2^7}\int^{\infty}_{0}e^{-x}\cdot x^6dx$

Now let $\displaystyle I(n) = \int^{\infty}_{0}e^{-nx}dx = \frac{1}{n}$

Then Differentiate $6$ times both side w r to $n$

$$I''''''(n) = \int^{\infty}_{0}e^{-nx}\cdot x^6dx = \frac{6!}{n^7}$$

so $$I = \frac{1}{2^7}\int^{\infty}_{0}e^{-x}\cdot x^6dx = \frac{6!}{2^7}$$

For $\displaystyle J = \int^{2}_{0}x(8-x^3)^{\frac{1}{3}}dx$ put $8-x^3=t^3,$

Then $x^2dx = -t^2dt$

so $$\displaystyle J = \int^{2}_{0}\frac{t}{(8-t^3)^{\frac{1}{3}}}dt$$

I did not know how to solve from here, could someone help me to solve it, thanks.

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DXT
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1 Answers1

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This kind of question requires knowledge of the Beta and Gamma functions and the relationship between them. In what follows, I use the Gamma function to compute $I$, the Beta function to compute $J$, and the latter's expression in terms of the Gamma function (as well as the Gamma function's reflection formula) to simplify the result. I recommend double-checking all my arithmetic, once you've learned the relevant material.

We have $I=\frac{\Gamma(7)}{2^7}=\frac{45}{8}$ and (by substituting $y=\frac{x^3}{8}$) $J=\frac{8}{3}\operatorname{B}(\frac{2}{3},\,\frac{4}{3})$ so $$IJ=15\operatorname{B}\left(\frac{2}{3},\,\frac{4}{3}\right)=5\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)=5\pi\csc\frac{\pi}{3}=\frac{10\pi}{\sqrt{3}}.$$

J.G.
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