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Claude Leibovici's user avatar
Claude Leibovici's user avatar
Claude Leibovici's user avatar
Claude Leibovici
  • Member for 11 years, 1 month
  • Last seen this week
  • Pau, France
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awarded
 Nice Answer
comment
Possible new series for $\pi$
@PeterTaylor. Thanks. What is interesting is thé diagonal part makes "most" of thé job. Cheers
comment
Possible new series for $\pi$
$$\frac{2x \sqrt{1-x^2} +8 \sin ^{-1}\left(\frac{\sqrt{1-\sqrt{1-x^2 }}}{\sqrt{2}}\right)-2 \sin ^{-1}(x)}{x}$$
comment
Possible new series for $\pi$
Did you find any other (even as some bizarre hypergeometric functions) ?
comment
Can this integral be solved analytically
@GeraldEdgar. $$\int \frac{dt}{e^t+e^{a t}}=-e^{-t} \, _2F_1\left(1,\frac{1}{1-a};\frac{a-2}{a-1};-e^{(a-1) t}\right)$$
awarded
 Yearling
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Solving 'impossible' integrals with a new (?) trick
@EmmanuelJoséGarcía. WA gives $0.298057$
revised
Solving 'impossible' integrals with a new (?) trick
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comment
Solving 'impossible' integrals with a new (?) trick
@EmmanuelJoséGarcía For the definite integral the result is $\frac{4}{3}+\sqrt{2}-\sqrt{6}$
comment
Solving 'impossible' integrals with a new (?) trick
@EmmanuelJoséGarcía. I shall add this one in my answer since it is nice.
answered
limit of definite integral as $N \to \infty$
comment
Solving 'impossible' integrals with a new (?) trick
$(6)$ is not the same here and on MSE
revised
Solving 'impossible' integrals with a new (?) trick
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revised
Solving 'impossible' integrals with a new (?) trick
added 518 characters in body
comment
Solving 'impossible' integrals with a new (?) trick
@EmmanuelJoséGarcía. You are correct. I shall fix it. I took it on MSE
answered
Solving 'impossible' integrals with a new (?) trick
awarded
 Enthusiast
revised
Zeroes of elementary polynomials without involving closed-form solutions
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revised
Zeroes of elementary polynomials without involving closed-form solutions
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answered
Zeroes of elementary polynomials without involving closed-form solutions
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