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? $\endgroup$ – NicoDean Oct 5 '14 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. $\endgroup$ – Sangchul Lee Oct 5 '14 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 Mar 21 '16 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. $\endgroup$ – Nicholas Stull Apr 12 '16 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. $\endgroup$ – Sangchul Lee May 25 '16 at 23:04

The following is only a partial answer, but it might be useful.

Assuming that all the parameters are positive, the integral $$ I(p,q,r) = \int_{0}^{1} \frac{\operatorname{arccot} q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \, dx $$ can be expressed in terms of $I \left(\frac{1}{q}, \frac{1}{p}, \frac{1}{r} \right)$.

$$ \small \begin{align} & \int_{0}^{1} \frac{\operatorname{arccot} q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \, dx \\ &= \int_{0}^{1} \int_{0}^{1} \frac{1}{t^{2}+p^{2}q^{2}x^{2}+q^{2}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dt \, dx \\&= \int_{0}^{1} \frac{(r^{2}+1)pqr}{q^{2}r^{2}+(r^{2}+1)t^{2}} \int_{0}^{1} \frac{1}{(r^{2}+1)p^{2}x^{2}+1} \, dx \, dt - \int_{0}^{1} \int_{0}^{1} \frac{pq^{3}r}{q^{2}r^{2}+(r^{2}+1)t^{2}} \frac{1}{t^{2}+p^{2}q^{2}x^{2}+q^{2}} \,dx \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - \int_{0}^{1} \frac{pq^{3}r}{q^{2}r^{2}+(r^{2}+1)t^{2}} \frac{\operatorname{arccot} \left(\frac{1}{p}\sqrt{\frac{t^{2}}{q^{2}}+1} \right)}{pq^{2}\sqrt{\frac{t^{2}}{q^{2}}+1}} \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - \int_{0}^{1} \frac{\frac{1}{pqr}}{\frac{1+r^{2}}{r^{2}} \frac{t^{2}}{q^{2}}+1}\frac{\operatorname{arccot} \left(\frac{1}{p}\sqrt{\frac{t^{2}}{q^{2}}+1} \right)}{p\sqrt{\frac{t^{2}}{q^{2}}+1}} \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - I \left(\frac{1}{q}, \frac{1}{p}, \frac{1}{r} \right).\end{align}$$

And by making the substitution $u= \frac{1}{x}$ followed by the substitution $w^{2}= p^{2}+u^{2}$, one can show that

$$ \begin{align} &\int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dx \\ &= \frac{\pi}{2} \int_{0}^{1} \frac{1}{ q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} dx - \int_{0}^{1} \frac{\text{arccot} \, q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dx \\ &= \frac{\pi}{2} \, \text{arctan} \left(\frac{pr}{\sqrt{p^{2}+1}} \right) -I(p,q,r). \end{align}$$

  • $\begingroup$ This integral is an Ahmed's integral only for $a=1$ and is equal to $A(1/2,2,1)$. $\endgroup$ – alexjo Nov 25 '13 at 16:49
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    $\begingroup$ Nice calculation. You can also generalize this calculation to relate the ahmed integral of parameter $(p, q, r)$ and that of $(1/q, 1/p, 1/r)$. $\endgroup$ – Sangchul Lee Nov 28 '13 at 11:49

I have written an article on arxiv on the relation between the Probability integral and Ahmed integral I have not the Url at hand by now but you can search through Google Scholar by the argument ''Probability integral and Ahmed integral''. It might suggest a new way to generalize the Ahmed Integral juan PLA

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    $\begingroup$ You mean the following? arxiv.org/pdf/1505.03314v2.pdf $\endgroup$ – b00n heT Jan 23 '17 at 17:39
  • $\begingroup$ Yes I mean this paper: arxiv.org/pdf/1505.03314v2.pdf $\endgroup$ – PLA Jan 24 '17 at 21:34
  • $\begingroup$ Borwein and Bailey have also tried to generalize Ahmed's integral in a note called ''Experimental Mathematics'' Can be retrieved through Google with this title and the names of the authors $\endgroup$ – PLA Jan 24 '17 at 21:38
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    $\begingroup$ Not to be rude, but how is this an answer? $\endgroup$ – The Count Feb 2 '17 at 3:03
  • $\begingroup$ @Pla: your proof is amazing and nice (too bad your first article about the method you use is not available on arxiv). J'adore vos deux articles sur le sujet ! $\endgroup$ – FDP Apr 14 '17 at 14:44

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