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May
1
comment Compute $I = \int_0^{2\pi} \frac{ac-b^2}{[a \sin(t)+ b \cos(t)]^2+[b \cos(t)+ c \sin(t)]^2}dt$
@Thramo The formula printed in Courant is incorrect. Try plugging in values for $a,b,c$ into Courant's formula and comparing the LHS and RHS numerically. For example, setting $a=2,b=1,c=-2$ the integral on the LHS will equal $-\frac52\pi$, but the expression on the RHS is $-2\pi$.
May
1
answered Simplifying a certain polylogarithmic sum in two variables
May
1
revised Simplifying a certain polylogarithmic sum in two variables
modified question statement for notational consistency with posted response
May
1
answered Compute $I = \int_0^{2\pi} \frac{ac-b^2}{[a \sin(t)+ b \cos(t)]^2+[b \cos(t)+ c \sin(t)]^2}dt$
Apr
22
comment Integral of $\frac{x}{(1-x^3)\sqrt{1-x^2}}$
@Transcendental Although it would be a pain to verify, WRA's result is ultimately the same. The steps WRA uses weren't available, but based previous experience my guess is that it 1) starts by subbing $x=\sin{\theta}$ like you did, then 2) uses the tangent-half-angle substitution. The main difference now is that WRA's default assumption is that parameters can be any complex number, and it will use the full set of complex roots of the denominator to fully decompose the integrand by partial fractions. The end result is that WRA gets a lot of complex log terms instead of arctan ones.
Apr
22
answered Integral of $\frac{x}{(1-x^3)\sqrt{1-x^2}}$
Apr
22
comment Wolfram answer is different for the integral $\sqrt{\frac{x}{2-x}}dx$
The logarithmic term is actually the arcsin function in disguise because of complex number magic. Wolfram Alpha is programmed to think logarithms are 'simpler' than trig functions even if the resulting expresion is longer.
Apr
1
comment Some diffuculties trying to compute double sums
It would be helpful if you could give us more detail on what exactly is confusing you here. The more effort you put into your question, the more effort we put into our answers. =)
Mar
31
answered Is there a nice finite sum expression for $\frac{x^{2n}(1-x)^{2n}}{1+x^2}$?
Mar
31
comment 3-dimensional integral with a step function
@Elina I would also like to impress upon that I spent months searching in vain for a simpler solution. First I tried to reduce my Appell $F_3$ function to elliptic integrals and gave up. Then I tried to reduce it to single-variable (generalized) hypergeometric functions and gave up. Finally I tried to reduce it to Appell $F_1$ or $F_2$ functions (which are somewhat simpler than the $F_3$ functions, relatively speaking), and again I gave up. But while the general case doesn't seem simplify, some particular cases nevertheless do. It would be very interesting if your integral is one of these.
Mar
31
comment 3-dimensional integral with a step function
@Elina Sorry for taking so long getting back to you. Of the two integrals you mentioned, the first is of course an elliptic integral, making this integral in principle quite easy, though computing it by hand would involve tortuous amounts of algebra. The second integral is much more subtle, and belongs to a family of integrals I've been studying off and on for nearly a year. See my question here: math.stackexchange.com/questions/1300459/… . There I find a solution in terms of Appell $F_3$ functions.
Mar
21
answered 3-dimensional integral with a step function
Mar
20
answered Serendipitous mathematical discoveries in recent times
Mar
20
comment How can one solve the following equation for Mathcad cannot calculate that
@ТимофейЛомоносов The solution will depend on the value of the parameter, in particular $\gamma$.
Mar
20
comment 3-dimensional integral with a step function
@Elina I believe this integral may be solved exactly in terms of elliptic integrals in the case you are interested in. Looks tough but certainly doable. Out of curiosity, what exactly is the physical context of the problem you are trying to solve?
Mar
19
awarded  Custodian
Mar
16
awarded  Nice Answer
Mar
14
comment Integration using Euler $\int{\sqrt{x^2+2x-1} }/{x}\,dx$
@DraganZrilić Since this quadratic has two distinct real roots, Euler's third substitution will probably offer the cleanest solution.
Mar
6
revised Closed form double integral $ \int_{a}^{c}dr \int_{b}^{d} dr' \, \frac{r r'}{\sqrt{(r - a)(r' - b)(r-c)(r'-d)}} \frac{r_<^{\ell}}{r_>^{\ell+1}}$
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