The following integral

∫10arctan√x2+2√x2+2dxx2+1=5π296

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 integralof parameter p, q and r byA(p,q,r):=∫10arctanq√p2x2+1q√p2x2+1pqrdx(r2+1)p2x2+1.

Now suppose that pqr=1, and define its complementary parameters as

˜p=r√q2+1,˜q=p√r2+1,and˜r=q√p2+1,

Then my guess is that

A(p,q,r)=π28+12{arctan2(1/˜p)−arctan2(˜q)−arctan2(˜r)}.

Plugging the values (p,q,r)=(1/√2,√2,1), the corresponding complementary parameters become (˜p,˜q,˜r)=(√3,1,√3). Then for these choices, the original Ahmed’s integral (1) is retrieved:

∫10arctan√x2+2√x2+2dxx2+1=π28+12{arctan21√3−arctan21−arctan2√3}=5π296.

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 (2). Then whenever k≤1, we haveA(p,q,r)=2χ2(k)−karctan(˜p)arctan(k˜p)+k2∫1011−k2x2log(1+˜p2x21+˜p2×1+˜q2x21+˜q2×1+˜r2x21+˜r2)dx.

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.

**Answer**

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)=∫10arccotq√p2x2+1q√p2x2+1pqr(r2+1)p2x2+1dx can be expressed in terms of I(1q,1p,1r).

∫10arccotq√p2x2+1q√p2x2+1pqr(r2+1)p2x2+1dx=∫10∫101t2+p2q2x2+q2pqr(r2+1)p2x2+1dtdx=∫10(r2+1)pqrq2r2+(r2+1)t2∫101(r2+1)p2x2+1dxdt−∫10∫10pq3rq2r2+(r2+1)t21t2+p2q2x2+q2dxdt=(r2+1)pqrarctan(√r2+1qr)qr√r2+1arctan(p√r2+1)p√r2+1)−∫10pq3rq2r2+(r2+1)t2arccot(1p√t2q2+1)pq2√t2q2+1dt=(r2+1)pqrarctan(√r2+1qr)qr√r2+1arctan(p√r2+1)p√r2+1)−∫101pqr1+r2r2t2q2+1arccot(1p√t2q2+1)p√t2q2+1dt=(r2+1)pqrarctan(√r2+1qr)qr√r2+1arctan(p√r2+1)p√r2+1)−I(1q,1p,1r).

And by making the substitution u=1x followed by the substitution w2=p2+u2, one can show that

∫10arctanq√p2x2+1q√p2x2+1pqr(r2+1)p2x2+1dx=π2∫101q√p2x2+1pqr(r2+1)p2x2+1dx−∫10arccotq√p2x2+1q√p2x2+1pqr(r2+1)p2x2+1dx=π2arctan(pr√p2+1)−I(p,q,r).

**Attribution***Source : Link , Question Author : Sangchul Lee , Answer Author : Random Variable*