# Find the value of $\int_{0}^{\sqrt{3}}\int_{0}^{\tan^{-1}y}\sqrt{xy}\ dxdy$

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?

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

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.

Edit

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 $$I=0.69351943$$

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