How can I find the value of $I = \int_{0}^{\sqrt{3}}\int_{0}^{\tan^{-1}y}\sqrt{xy}\ dxdy$?

I tried changing the limits and got the new integral as $I = \int_{0}^{\pi/3}\int_{\tan(x)}^{\sqrt{3}}\sqrt{xy}\ dydx$

And then tried evaluating it to get $I = \frac{2}{3} \int_{0}^{\frac{\pi}{3}}\sqrt{x}(3^\frac{3}{4} - \tan^\frac{3}{2}x)\ dx$ but I do not know how to proceed further to get the result

Wolfram alpha gives the same answer on all cases so I don't think my steps are wrong. Or is this an XY problem at this point?

  • $\begingroup$ \begin{align}\int_0^\sqrt{3}\sqrt{y}\left(\int_0^{\arctan y}\sqrt{x}dx\right)dy\end{align} $\endgroup$
    – FDP
    Dec 9, 2022 at 20:02

1 Answer 1


Too long for a comment.

Your steps are very correct. The problem is that none of the CAS I used has been able to compute the second antiderivative or definite integral.

Using your second solution (which is better for the series solution approach) $$I = \frac{2}{3} \int_{0}^{\frac{\pi}{3}} \sqrt{x}\, \left(3^{3/4}-\tan ^{\frac{3}{2}}(x)\right)\, dx=\frac{4 \pi ^{3/2}}{9\ 3^{3/4}}- \frac{2}{3} \int_{0}^{\frac{\pi}{3}} \sqrt{x}\,\tan ^{\frac{3}{2}}(x)\, dx$$ Expanding as a series built around $x=0$ $$ \sqrt{x}\,\tan ^{\frac{3}{2}}(x)=\sum_{n=1}^\infty a_n\,x^{2n}$$ where the first coefficients (given by Wolfram Alpha) are $$\left\{1,\frac{1}{2},\frac{29}{120},\frac{1693}{15120},\frac{4361}{8 6400},\frac{1321}{59136},\frac{6380159677}{653837184000},\cdots\right\}$$ give $$I=\frac{4 \pi ^{3/2}}{9\ 3^{3/4}}- \frac{2}{3}\sum_{n=1}^\infty\frac {a_n}{2n+1}\left(\frac{\pi }{3}\right)^{2 n+1}$$ Computing $$J_k=\frac{4 \pi ^{3/2}}{9\ 3^{3/4}}- \frac{2}{3}\sum_{n=1}^k \frac {a_n}{2n+1}\left(\frac{\pi }{3}\right)^{2 n+1}$$ Using only the coefficients given in the linked page $J_{10}=0.693568$ while numerical integration gives $I=0.693519$.

For sure, using more terms leads to better and better results.


Since we have the series expansion, we can easily build the $[2(n+1),2n]$ Padé approximant $P_n$ around $x=0$. They write $$P_n=x^2 \, \frac{1+\sum _{k=1}^n \alpha_k\, x^{2 k}}{1+\sum _{k=1}^n \beta_k\, x^{2 k} }$$ whose errors are $O(x^{4(n+1)})$.

Just to give an idea, consider the norm $$\Phi_n=\int_0^{\frac \pi 3}\Big[\sqrt{x}\,\tan ^{\frac{3}{2}}(x)-P_n\Big]^2\,dx$$ Starting from $n=1$, they are $$\{7.60\times 10^{-5},3.45\times 10^{-10},1.95\times 10^{-11},4.77\times 10^{-16},2.40\times 10^{-21}\}$$

After factor decomposition, wa have $$P_n=\sum_{k=1}^n A_k \frac {x^2}{x^2-r_k}\qquad \implies \qquad \int_{0}^{\frac{\pi}{3}}P_n\,dx=\sum_{k=1}^n A_k \left(\frac{\pi }{3}-\sqrt{r} \tanh ^{-1}\left(\frac{\pi }{3 \sqrt{r_k}}\right)\right)$$

Up to $n=4$, we have (with readicals) the exact values of the $r_k$. So, considering the simple $P_3$ for which the $\alpha_k$ are $$\left\{-\frac{4307737}{124259668},-\frac{6720869017}{139419347496 },\frac{15215045947}{13012472432960}\right\}$$ and the $\beta_k$ are $$\left\{-\frac{66437571}{124259668},-\frac{15712003321}{6970967374 80},\frac{10427397914831}{351336755689920}\right\}$$

Converted to decimals

$$\int_0^{\frac \pi 3} P_3\,dx=0.58824037\qquad \implies \qquad \color{red}{I=0.69351}853$$ while numerical integration leads to $I=0.69351943$

  • $\begingroup$ Ooh, so near to $\ln2\approx0.693147$ and yet so far. $\endgroup$
    – J.G.
    Dec 10, 2022 at 6:42
  • $\begingroup$ @J.G. Not so close. But $\frac{5 \left(\sqrt{2+\sqrt{3}}-1\right)}{4+e}$ is in error of $10^{-7}$. Cheers :-) $\endgroup$ Dec 10, 2022 at 6:51

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