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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.

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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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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

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

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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