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The following integral

\begin{align*} \int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{dx}{x^{2}+1} = \frac{5\pi^{2}}{96} \tag{1} \end{align*}

is called the Ahmed's integral and became famous since its first discovery in 2002. Fascinated by this unbelievable closed form, I have been trying to generalize this result for many years, though not successful so far.

But suddenly it came to me that some degree of generalization may be possible. My conjecture is as follows: Define the (generalized) Ahmed integral of parameter $p$, $q$ and $r$ by

\begin{align*} A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1}}{q \sqrt{p^{2}x^{2} + 1} } \, \frac{pqr \, dx}{(r^{2} + 1)p^{2} x^{2} + 1}. \end{align*}

Now suppose that $p q r = 1$, and define its complementary parameters as

\begin{align*} \tilde{p} = r \sqrt{\smash{q}^{2} + 1}, \quad \tilde{q} = p \sqrt{\smash{r}^{2} + 1}, \quad \text{and} \quad \tilde{r} = q \sqrt{\smash{p}^{2} + 1}, \tag{2} \end{align*}

Then my guess is that

\begin{align*} A(p, q, r) = \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} (1 / \tilde{p} ) - \arctan^{2} ( \tilde{q} ) - \arctan^{2} ( \tilde{r} ) \right\}. \end{align*}

Plugging the values $(p, q, r) = (1/\sqrt{2}, \sqrt{2}, 1)$, the corresponding complementary parameters become $(\tilde{p}, \tilde{q}, \tilde{r}) = (\sqrt{3}, 1, \sqrt{3})$. Then for these choices, the original Ahmed's integral $\text{(1)}$ is retrieved:

\begin{align*} \int_{0}^{1} \frac{\arctan \sqrt{x^{2} + 2} }{\sqrt{x^{2} + 2} } \, \frac{dx}{x^{2} + 1} &= \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} \frac{1}{\sqrt{3}} - \arctan^{2} 1 - \arctan^{2} \sqrt{3} \right\} \\ &= \frac{5\pi^{2}}{96}. \end{align*}

In fact, I have a more generalized conjecture involving dilogarithms depending on complementary parameters. But since this specialized version is sufficiently daunting, I won't deal with it here.

Unfortunately, proving this relation is not successful so far. I just heuristically calculated and made some ansatz to reach this form. Can you help me improve the situation by proving this or providing references to some known results?


EDIT. I finally succeeded in proving a general formula: let $k = pqr$ and complementary parameters as in $\text{(2)}$. Then whenever $k \leq 1$, we have

\begin{align*} A(p, q, r) &= 2\chi_{2}(k) - k \arctan (\tilde{p}) \arctan \left( \frac{k}{\tilde{p}} \right) \\ &\quad + \frac{k}{2} \int_{0}^{1} \frac{1}{1-k^{2}x^{2}} \log\left( \frac{1+\tilde{p}^{2}x^{2}}{1+\tilde{p}^{2}} \times \frac{1+\tilde{q}^{2}x^{2}}{1+\tilde{q}^{2}} \times \frac{1+\tilde{r}^{2}x^{2}}{1+\tilde{r}^{2}} \right) \, dx. \end{align*}

Then the proposed conjecture follows as a corollary. I'm planning to gather materials related to the Ahmed's integrals and put into a combined one. You can find an ongoing proof of this formula here.

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    $\begingroup$ Sorry for the off-topic question: I have looked through some of your very beautiful answers here, and it seems you are developing cool new methods. Are you publishing those methods/generalisations in some math-journals, or are they on Math.SE only? $\endgroup$ Commented Oct 5, 2014 at 17:34
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    $\begingroup$ @NicoDean, Thank you! I think that most of the techniques are already well established, so I am only posting them on Math.SE. $\endgroup$ Commented Oct 5, 2014 at 22:02
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    $\begingroup$ This is a great work! I've just met Ahmed's integral (and it's solution) in the book "Inside interesting integrals...", I had no idea it's possible to generalize it to such extent $\endgroup$
    – Yuriy S
    Commented Mar 21, 2016 at 8:10
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    $\begingroup$ This might be one of the more beautiful things I've seen, but I'm going to have to read over that proof 3 or 4 times to fully appreciate this. $\endgroup$ Commented Apr 12, 2016 at 16:09
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    $\begingroup$ Thank you so much for all these responses. My plan is to polish the result before I make any edit on this question, but unfortunately, I am recently so busy with my own study to improve it. And moreover on this, I suspect that my computation is essentially a disguise of Lobachevski's volume formula for an orthoscheme (given the relationship between the generalized Ahmed's integral and the Coxeter's integral). I would like to clarify this linkage before I make any update. $\endgroup$ Commented May 25, 2016 at 23:04

2 Answers 2

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Below is a simpler, more direct proof of the proposed conjecture. First, establish

\begin{align*} K(a)=\int_{0}^{1} \frac{\ln\frac{a+x^2}{a+1}}{1-x^2} dx =&\int_0^1 \int_{\infty}^a\frac1{(y+1)(y+x^2)}dy\ dx\\ =&\int_\infty^a \frac{\text{arccot} \sqrt {y}}{\sqrt{y}(y+1)}dy= -\text{arccot}^2 \sqrt {a} \end{align*} Then, for $r=\frac1{pq}$ \begin{align*} & \int_{0}^{1} \frac{1}{(p^{2} + \frac1{q^2})x^{2} + 1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1}}{q \sqrt{p^{2}x^{2} + 1} } \ dx\\ =& \int_0^1 \frac{q^2}{(p^{2}q^2 + 1)x^{2} + q^2} \int_0^x \frac{x}{x^2+q^2(p^2x^2+1)y^2}dy\ dx\\ =&\int_0^1 \int_y^1\frac{q^2x}{[(p^2q^2+1)x^2+q^2][(1+p^2q^2y^2)x^2+q^2y^2]}dx\ dy\\ =&\ \frac12 \int_0^1 \frac1{1-y^2}\ln\frac{[1+(p^2+1)q^2y^2][q^2+(1+p^2q^2)y^2]}{y^2[1+q^2+p^2q^2y^2] [1+q^2 +p^2q^2]} \ dy\\ =&\ \frac12\int_0^1 \frac1{1-y^2}\left[-\ln y^2 -\ln \frac{1+q^2+p^2q^2y^2}{1+q^2 +p^2q^2}\right.\\ &\hspace{25mm}\left.+ \ln \frac{1+(1+p^2)q^2y^2}{1+q^2 +p^2q^2} + \ln \frac{q^2+(1+p^2q^2)y^2}{1+q^2 +p^2q^2}\right] dy\\ =&\ \frac12\left[-K(0)-K\left(\frac{q^2+1}{p^2q^2} \right) + K\left(\frac{1}{p^2q^2+q^2} \right)+ K\left(\frac{q^2}{p^2q^2+1} \right) \right]\\ =&\ \frac{\pi^2}8+\frac12\bigg[ \arctan^2\frac{pq}{\sqrt{q^2+1}}-\arctan^2 q\sqrt{p^2+1} -\arctan^2{\frac{ \sqrt{p^2q^2+1}}q}\bigg] \end{align*}

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    $\begingroup$ Genius +1, thank you for sharing. $\endgroup$
    – OnTheWay
    Commented Jan 27 at 15:45
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Too long for comment ,Define the generalized Ahmed integral $$I\left( p,q,r \right) = pqr \int_0^1{\frac{\arctan \left( q\sqrt{p^2u^2+1} \right)}{q\sqrt{p^2u^2+1}}\frac{1}{\left( p^2\left( r^2+1 \right) u^2+1 \right)}}\,du $$

$ 1 \text{, when } p,q,r>0, \, pqr\le 1 \text{, then: } $ $$ \begin{aligned} &I\left( p,q,r \right) = pqr\int_0^1{\frac{\arctan \left( q\sqrt{p^2u^2+1} \right)}{q\sqrt{p^2u^2+1}}\frac{1}{\left( p^2\left( r^2+1 \right) u^2+1 \right)}}\,du \\ &=A\left( p,q,r \right) -\arctan \left( \frac{pq}{\sqrt{q^2+1}} \right) \arctan \left( r\sqrt{q^2+1} \right) \\ &A\left( p,q,r \right) =2\sum_{k=0}^{\infty}{\frac{\left( pqr \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}+ \\ &\frac{pqr}{2}\int_0^1{\frac{\ln \left( \frac{1+\left( r\sqrt{q^2+1} \right) ^2x^2}{1+\left( r\sqrt{q^2+1} \right) ^2} \right) +\ln \left( \frac{1+\left( p\sqrt{r^2+1} \right) ^2x^2}{1+\left( p\sqrt{r^2+1} \right) ^2} \right) +\ln \left( \frac{1+\left( q\sqrt{p^2+1} \right) ^2x^2}{1+\left( q\sqrt{p^2+1} \right) ^2} \right)}{1-\left( pqr \right) ^2x^2}dx} \end{aligned} $$

$ 2 \text{, for any } p,q,r>0, \text{ then: } $ $$ \begin{aligned} &I\left( p,q,r \right) = pqr\int_0^1{\frac{\arctan \left( q\sqrt{p^2u^2+1} \right)}{q\sqrt{p^2u^2+1}}\frac{1}{\left( p^2\left( r^2+1 \right) u^2+1 \right)}}\,du \\ &=\hat{A}\left( p,q,r \right) -\arctan \left( \frac{pq}{\sqrt{q^2+1}} \right) \arctan \left( r\sqrt{q^2+1} \right) \\ &\hat{A}\left( p,q,r \right) =\frac{\pi ^2}{4}+2\sum_{k=0}^{\infty}{\frac{\left( \frac{pqr-1}{pqr+1} \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}+ \\ &\frac{pqr}{2}\int_0^1{\frac{\ln \left( \frac{1+\left( r\sqrt{q^2+1} \right) ^2x^2}{1+\left( r\sqrt{q^2+1} \right) ^2}\frac{1+\left( p\sqrt{r^2+1} \right) ^2x^2}{1+\left( p\sqrt{r^2+1} \right) ^2}\frac{1+\left( q\sqrt{p^2+1} \right) ^2x^2}{1+\left( q\sqrt{p^2+1} \right) ^2} \right) +4\ln \left( pqr \right)}{1-\left( pqr \right) ^2x^2}dx} \end{aligned} $$

For this problem, note that

$$ I\left( p,q,r \right) =\int_0^1{\frac{\arctan \left( q\sqrt{p^2u^2+1} \right)}{q\sqrt{p^2u^2+1}}\frac{1}{\left( p^2\left( r^2+1 \right) u^2+1 \right)}du}=\int_0^1{\frac{\arctan \left( \frac{2\sqrt{2u^2+1}}{\sqrt{5}\left( u^2+1 \right)} \right)}{\sqrt{2u^2+1}\left( 3u^2+1 \right)}du}, \\ \text{using the addition formula for } \arctan \left( x \right): \arctan \left( \frac{2\sqrt{2u^2+1}}{\sqrt{5}\left( u^2+1 \right)} \right) \\ =\arctan \left( \sqrt{5}\sqrt{2u^2+1} \right) -\arctan \left( \frac{\sqrt{2u^2+1}}{\sqrt{5}} \right) \text{; so } I=\int_0^1{\frac{\arctan \left( \sqrt{5}\sqrt{2u^2+1} \right) -\arctan \left( \frac{\sqrt{2u^2+1}}{\sqrt{5}} \right)}{\sqrt{2u^2+1}\left( 3u^2+1 \right)}du} \\ =I\left( \sqrt{2},\sqrt{5},\frac{1}{\sqrt{2}} \right) -I\left( \sqrt{2},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{2}} \right) $$

where $$I\left( \sqrt{2},\sqrt{5},\frac{1}{\sqrt{2}} \right) = 2\sum_{k=0}^{\infty}{\frac{\left( \frac{\sqrt{5}-1}{\sqrt{5}+1} \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}+ \\ \int_0^1{\frac{\sqrt{5}\left( \ln \left( \frac{\left( \frac{\sqrt{\sqrt{5}^2+1}x}{\sqrt{2}} \right) ^2+1}{\left( \frac{\sqrt{\sqrt{5}^2+1}}{\sqrt{2}} \right) ^2+1}\frac{\left( \frac{\sqrt{\sqrt{2}^2+1}x}{\frac{1}{\sqrt{5}}} \right) ^2+1}{\left( \frac{\sqrt{\sqrt{2}^2+1}}{\frac{1}{\sqrt{5}}} \right) ^2+1}\frac{\left( \sqrt{2}\sqrt{\left( \frac{1}{\sqrt{2}} \right) ^2+1}x \right) ^2+1}{\left( \sqrt{2}\sqrt{\left( \frac{1}{\sqrt{2}} \right) ^2+1} \right) ^2+1} \right) +4\ln \left( \sqrt{5} \right) \right)}{2\left( 1-\left( \sqrt{5}x \right) ^2 \right)}}\,dx+ \\ \frac{\pi ^2}{4}-\arctan \left( \frac{\sqrt{\sqrt{5}^2+1}}{\sqrt{2}} \right) \arctan \left( \frac{\sqrt{2}\sqrt{5}}{\sqrt{\sqrt{5}^2+1}} \right) \\ =\frac{2\left( \frac{1}{2}\sqrt{5}\ln\left(1+\sqrt{2}\right) -\frac{1}{2}\sqrt{5}\ln\left(1-\sqrt{2}\right) \right)}{\sqrt{5}}+ \\ \int_0^1{\frac{\sqrt{5}\left( 2\ln \left( 3x^2+1 \right) +\ln \left( \frac{25}{256}\left( 15x^2+1 \right) \right) \right)}{2\left( 1-\left( \sqrt{5}x \right) ^2 \right)}}\,dx \\ +\frac{1}{12}\pi \left( 3\pi -4\arctan \left( \sqrt{\frac{5}{3}} \right) \right); $$

Similarly, we get $$I\left( \sqrt{2},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{2}} \right) =2\sum_{k=0}^{\infty}{\frac{\left( \frac{1}{\sqrt{5}} \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}+ \\ \int_0^1{\frac{2\ln \left( \frac{5}{8}\left( \frac{3x^2}{5}+1 \right) \right) +\ln \left( \frac{1}{4}\left( 3x^2+1 \right) \right)}{2\sqrt{5}\left( 1-\frac{x^2}{5} \right)}}\,dx-\arctan \left( \frac{\sqrt{2}}{\sqrt{5}\sqrt{\left( \frac{1}{\sqrt{5}} \right) ^2+1}} \right) \arctan \left( \frac{\sqrt{\left( \frac{1}{\sqrt{5}} \right) ^2+1}}{\sqrt{2}} \right) \\ =\frac{2\left( \frac{1}{2}\sqrt{5}\ln\left(1+\sqrt{\frac{5}{2}}\right) -\frac{1}{2}\sqrt{5}\ln\left(1-\sqrt{\frac{5}{2}}\right) \right)}{\sqrt{5}}+ \\ \int_0^1{\frac{2\ln \left( \frac{5}{8}\left( \frac{3x^2}{5}+1 \right) \right) +\ln \left( \frac{1}{4}\left( 3x^2+1 \right) \right)}{2\sqrt{5}\left( 1-\frac{x^2}{5} \right)}}\,dx-\frac{1}{6}\pi \tan ^{-1}\left( \sqrt{\frac{3}{5}} \right); $$

Then the integral is given by $$ \int_0^1{\frac{2\ln \left( \frac{5}{8}\left( \frac{3x^2}{5}+1 \right) \right) +\ln \left( \frac{1}{4}\left( 3x^2+1 \right) \right)}{2\sqrt{5}\left( 1-\frac{x^2}{5} \right)}dx}, \\ \int_0^1{\frac{\sqrt{5}\left( 2\ln \left( 3x^2+1 \right) +\ln \left( \frac{25}{256}\left( 15x^2+1 \right) \right) \right)}{2\left( 1-\left( \sqrt{5}x \right) ^2 \right)}}\,dx $$ an be computed accordingly

Regarding Sangchul Lee's paper "Some properties on generalized Ahmed's integral," two corrections are made (as amended in this document)

(1) Formula (1,3) $$\arctan \left( \frac{pq}{\sqrt{q^2+1}} \right) \arctan \left( r\sqrt{p^2+1} \right)$ should be corrected to $\arctan \left( \frac{pq}{\sqrt{q^2+1}} \right) \arctan \left( r\sqrt{q^2+1} \right)$$

(2) In formula (1,4), $$\chi_2(z)=\sum_{k=0}^{\infty}{\frac{\left(z \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}$$, the subscript should start from 0

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