Ahmed integrals usually contain unconventional factors. The initial one is
$$
\int_{0}^{1} \frac{\arctan\left ( \sqrt{2+x^2} \right ) }{
\left ( 1+x^2 \right )\sqrt{2+x^2} }\text{d}x
=\frac{5\pi^2}{96},
$$
which can be easily derived by double integration. In fact, such integrals are closely related to $K$-theory determining the algebraic relations among homogeneous polylogarithm values (e.g. polylogarithm ladders). Nevertheless, this is just for single $\arctan$ cases(but already has great complexity). Here gives a high-dimensional generalization differing from usual Ahmed integrals out of its nasty appearance. Now, we may verify that virtually every Ahmed-like integral, including high-dimensional generalizations, could be written into a linear combination of several integrals of error function. Namely, we have,
$$
I\in\left<\int_{0}^{\infty}e^{-ax}\prod_{i\in I}\text{erf}(b_ix)^{c_i}\text{d}x
\right>,
$$
where we also include regularized values (e.g. $a\le0$ when the integrals don't exist).
When we interpret it for Bessel functions, we have Bessel moments, from which we can derive some analogues that act similarly to Ahmed integrals but are much less obvious.
Define complete elliptic integrals of the first and second kind
respectively,$$
K(k)=\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-k^2t^2} }\text{d}t,
\quad E(k)=\int_{0}^{1} \frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2} }\text{d}t
$$
where $k$ is an elliptic modulus. And $K_0(z)$ represents the modified Bessel function of the second kind with order zero, and we use its integral representation:
$$
K_0(x)=\int_{1}^{\infty} \frac{e^{-tx}}{\sqrt{t^2-1} }
\text{d}t.
$$
$I_0(x)$ the first kind,
$$
I_0(x)=\frac{1}{\pi} \int_{-1}^{1}
\frac{e^{-tx}}{\sqrt{1-t^2} }\text{d}t,
$$
For single $K$:
- We have(asked from Wolfram Alpha) $$\int_{0}^{\infty} \frac{e^{-2x}}{\sqrt{x} }K_0(x)^2\text{d}x= 2\sqrt{2\pi} \cdot\frac{\Gamma\left ( \frac18 \right )^2 \Gamma\left ( \frac38 \right )^2}{32\pi}.$$ Write it as a multiple integral, $$ \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{\sqrt{x^2-1}\sqrt{y^2-1}\sqrt{2+x+y} } \text{d}x\text{d}y, $$ while noticing that for $a\ge1$, $ \int_{1}^{\infty} \frac{1}{\sqrt{x^2-1}\sqrt{a+x} } \text{d}x =\frac{2}{\sqrt{a+1} } K\left ( \sqrt{\frac{a-1}{a+1} } \right ). $ We can reformulate it as $$ \int_{0}^{1} \frac{1}{\sqrt{1-k^4} }K\left ( \sqrt{\frac{1+k^2}{2} } \right ) \text{d} k =\frac{\Gamma\left ( \frac18 \right )^2 \Gamma\left ( \frac38 \right )^2}{32\pi}. $$ This is manageable by other approaches.
- We have $$\int_{0}^{1}\frac{K\left ( \frac{1+x^2}{2} \right ) }{ \sqrt{1-x^2} }\text{d}x=\frac{1}{\pi\sqrt{2}}\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{1}{3-\cos x-\cos y-\cos z}\text{d}x\text{d}y\text{d}z\\ =\frac{\Gamma\left ( \frac{1}{24} \right ) \Gamma\left ( \frac{5}{24} \right)\Gamma\left ( \frac{7}{24} \right ) \Gamma\left ( \frac{11}{24}\right ) }{32\pi\sqrt{3}},$$ $$ \int_{0}^{1} \frac{K\left (\sqrt{ 1-\left(\frac{1-x^2}{2}\right)^2 } \right )}{\sqrt{1+x^2} } \text{d} x =\frac{\Gamma\left ( \frac{1}{24} \right ) \Gamma\left ( \frac{5}{24} \right)\Gamma\left ( \frac{7}{24} \right ) \Gamma\left ( \frac{11}{24}\right ) }{48\pi\sqrt{3}}, $$ which is equivalent to $\int_{0}^{\infty}e^{-3z}K_0(z)^2I_0(z)\text{d}z=\frac{\pi^2}3\int_{0}^{\infty}e^{-3z}I_0(z)^3\text{d}z$. And we also come to an extraordinary identity $$ \int_{0}^{1} \frac{K\left ( x \right ) }{\sqrt{3-x} } \text{d}x=\frac{\Gamma\left ( \frac{1}{24} \right ) \Gamma\left ( \frac{5}{24} \right)\Gamma\left ( \frac{7}{24} \right ) \Gamma\left ( \frac{11}{24}\right ) }{96\pi\sqrt{3}}. $$ (I had wished we could evaluate $\int_{0}^{1} \frac{K(k)}{\sqrt{4-k^2} } \text{d}k$.)
For double $K,E$:
- From here we have $$ \int_{0}^{1}K\left ( \sqrt{1-x^2} \right ) K\left ( \sqrt{1-x^4} \right )\text{d}x =\frac{1}{16}{\Gamma\left ( \frac14 \right )^4} {}_4F_3\left ( \frac14,\frac14,\frac14,\frac14; \frac12,\frac12,1;1 \right )-\frac14{\Gamma\left ( \frac34 \right )^4} {}_4F_3\left ( \frac34,\frac34,\frac34,\frac34; 1,\frac32,\frac32;1 \right ), $$ $$ \int_{0}^{1} k^2K\left (\sqrt{ 1-k^4 }\right ) \left [ \frac{K\left ( \sqrt{1-k^2} \right ) -E\left ( \sqrt{1-k^2} \right ) }{1-k^2} +\frac{2}{\sqrt{1+k^2} } E\left ( \frac{k}{\sqrt{1+k^2} } \right ) \right ] \text{d} k=\frac{\pi^2}{4}. $$
- Evaluating $\int_{0}^{\infty} \sqrt{z} e^{-3z}\left ( K_0(z)^3-3\pi^2K_0(z)I_0(z)^2\right ) \text{d} z=-\frac{\pi^2}{2\sqrt{2} } \sqrt{\pi} $ gives $$ \int_{0 }^{\sqrt{\frac23}} \frac{(K(k)-E(k))}{ k\sqrt{1-k^2} }K\left (\frac{k\sqrt{2-3k^2} }{1-k^2} \right )\text{d} k-\frac13 \int_{\sqrt{\frac23}}^1 \frac{\sqrt{1-k^2} (K(k)-E(k))}{ k(2k^2-1) }K\left (\frac{k\sqrt{3k^2-2} }{2k^2-1} \right )\text{d} k=\frac{\pi^2}{12}. $$
A related question: such integrals are actually periods on certain algebraic varieties. I am a bit confused about it. Any references to it?
Feel free to share your opinions and insights.
