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Martin.s
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Conjectured closed form of $\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy$
You didn't get it. I'm asking what I can learn from a response given by Mathematica software
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Conjectured closed form of $\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy$
This is closed form not an answer
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Conjectured closed form of $\int_0^1\frac{\ln^3(1+x)\,\ln^3x}x\mathrm{d}x$
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Conjectured closed form of $\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy$
answered
Structural differences between closed forms of two related infinite products?
accepted
Conjectured closed form of $\int_0^1 \frac{\text{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx$
comment
Conjectured closed form of $\int_0^1 \frac{\text{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx$
I'm not sure because the integral came up in my discussion group, and the result matches numerically.
comment
Conjectured closed form of $\int_0^1 \frac{\text{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx$
@GHfromMO sorry sir, I'm a new user on MathOverflow, and I edited the question, but I'm not sure if it was a good edit.
revised
Conjectured closed form of $\int_0^1 \frac{\text{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx$
deleted 117 characters in body; edited tags
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Conjectured closed form of $\int_0^1 \frac{\text{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx$
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comment
Conjectured closed form of $\int_0^1 \frac{\text{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx$
Why downvotes? Can someone explain
asked
Conjectured closed form of $\int_0^1 \frac{\text{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx$
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Positivity of the Fourier transform: prove or disprove that $\operatorname{Re}(\overline{\widehat{u}}(\xi) \widehat{F\circ u}(\xi))\geq0$
I don't see any reason why you would have the decay. Note that here, $|\widehat{u(t)}|^2\le c|\widehat{u(0}|^2$ doesn't imply decay, as $c$ may be greater than $1$. It just implies a kind of stability for the Fourier transform
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