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Jul
24
comment A possible dilogarithm identity?
Largely just aesthetics. It's slightly less clutterly, and slightly faster the type out. Unless there's a point later in the derivation where its value $\pi^2/6$ offers some non-trivial simplification, writing zeta just feels cleaner.
Jul
24
awarded  Revival
Jul
24
comment Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$
@VladimirReshetnikov Note that the integrals we need to calculate correspond to plugging in the values $a=\sqrt{2}$ and $b=\pm1$ into the general expressions I derived. When you go to compute the arguments of all the dilog terms required, you'll find that they're all expressible in terms of the silver ratio or its negative/reciprocal/etc., so it's not surprising most of the dilog terms can be eliminated. Also, see my response here.
Jul
24
answered Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$
Jul
24
comment Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$
@Zach466920 Actually, it's been my consistent experience with integrals like these that quite the opposite is the case. Wolfram Alpha in particualr can be quite bad at recognizing when massive cancellation is possible among even just a moderate handful of polylog terms. The most reliable way of obtaining the most compact final values tends to be by hand.
Jul
23
comment Definite integration.
@yasir They aren't equal. But their integrals from $0$ to $1$ are equal. This is shown by integrating by parts.
Jul
19
comment How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$
@user2468338 Since the quintic $16s^5-20s^3+5s-1$ is a polynomial with integer coefficients, we may use the rational root test to quickly determine that any rational roots must be powers of $\frac12$ up to $\frac{1}{16}$. Plugging $s=\frac{1}{2^n}$ we see that $s=1$ is the only rational root. After that, there are standard methods for factoring quartics.
Jul
13
comment Computing $\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}+1\right)}{k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$ in closed form
Impressive +1!) I really wasn't expecting this one to be solved because of how similar it is to this [long standing MSE question](math.stackexchange.com/questions/937912/… (one of my favorite problems of this entire site actually). Do you think your methods might shed any new light on that problem?
Jul
13
answered A possible dilogarithm identity?
Jul
13
comment Closed-form expression for an iterated integral
Thanks. In case you're curious, the key insight which made me think of the substitutions is the algebraic identity $(1+ab+ac+bc)\pm(a+b+c+abc)=(1\pm a)(1\pm b)(1\pm c)$. I encountered the identity while working on another problem and immediately thought of your integral here. It's almost painfully obvious in hindsight. sigh
Jul
12
awarded  Revival
Jul
12
answered Closed-form expression for an iterated integral
Jul
12
answered Definite integration.
Jul
9
comment Solving quartic equation using substitution
See Tschirnhaus transformations. Given that this problem was posed to you in the context of math history, I suspect this was the substitution the author of the problem had in mind. But Andre's suggestion is more straightforward I think.
Jul
9
comment How to solve this equation for x
@FooBar The real-analysis tag is appropriate, and maybe the special-functions tag would be too. I'm pretty sure you can write the solution to your equation in terms of the Lambert-W function.
Jul
9
comment How to solve this equation for x
This isn't actually a polynomial equation, except for the special cases where $\alpha$ is an integer. For general real-valued exponents, the power function is defined through the exponential function, which is transcendental.
Jul
8
comment Arclength of intersection between 2 perpendicular cylinders.
@CharlesBrillon Sure, that should be a straightforward generalization once you've digested the derivation above. Just apply the exact same transformations starting from an integral with arbitrary upper limit. The result is an incomplete elliptic integral like the one I evaluated in the appendix.
Jul
8
comment Arclength of intersection between 2 perpendicular cylinders.
@tired As it turns out, you don't even need elliptic integrals for the special case of equal radii. This special case may be handled with beta function machinery.
Jul
8
answered Arclength of intersection between 2 perpendicular cylinders.
Jul
2
comment The quadratic and cubic versions of a tough intregral
@Chris'ssistheartist Thanks. The pair of quadrilog terms are still bugging me though. I have a hunch they can be simplified to $Li_4(1/2)$ similar to the lower order terms, but I'm not sure how to show it without a corresponding Landen identity...