Ask Your Question
0

Problem with integrating the expression of M

asked 2019-07-22 10:43:22 +0200

Sha gravatar image

Hi. I have the following code that I want to integrate and differentiate, but I am stuck at the expression of M, where it is unable to integrate. Is my coding wrong?

c,t = var('c t')
Pi = RR.pi()
G=integrate(sqrt(1-t^2)*(t+c),t,-0.9,0.9);G
H=G.diff(c);H
L=integrate(-(t+c)/(sqrt(1-t^2)),t,-0.9,0.9);L
M=integrate(1/(sqrt(1-t^2)*(t-c)),t,-0.9,0.9);M #cannot seem to integrate this wrt to t
I=(c^2-1)*(L+(1-c^2)*M);I #equation I that involves M
P=I.diff(c);P #differentiate I wrt to c to obtain equation P

I have tried integrating M by hand which gives me a closed-form involving log function, but I can't seem to integrate it here using Sage.

edit retag flag offensive close merge delete

Comments

What do you assume about $c$? Try making it explicit, using assume before integrate.

rburing gravatar imagerburing ( 2019-07-22 14:29:14 +0200 )edit

Homework ?

Works for me using Sage 8.9.beta3 using Python3. rburning's hint is good, but trying other integrators ("sympy", "giac", "mathematica_free") might give you other ideas... Check your results by rederiving...

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2019-07-22 23:31:47 +0200 )edit

@rburing c is any value in the interval [-0.9, 0.9], which i plan to vary later at the end after the integration and differentiation.

Sha gravatar imageSha ( 2019-07-23 07:26:45 +0200 )edit

@emmanuel definitely not homework but research paper that I am leading at the moment. my co-author got different results than mine. so we thought of using Sage to verify the results.

Sha gravatar imageSha ( 2019-07-23 07:28:17 +0200 )edit

1 Answer

Sort by ยป oldest newest most voted
0

answered 2019-07-24 15:17:59 +0200

Emmanuel Charpentier gravatar image

updated 2019-07-24 16:06:02 +0200

Well, in this case, let's look at your expression's antiderivative (Sage is known to have some infelicities in definite integration). Sage's default integrator (i. e. maxima) needs to know if $c^2-1>0$ ; account for that as well as for other Sage's integrators :

(c,t)=var("c, t")
E=sqrt(1-t^2)*(t+c)
Sols={}
with assuming(c>-1,c<1):Sols.update({"Mn":E.integrate(t)})
with assuming(c<-1):Sols.update({"Mp":E.integrate(t)})
Sols.update({"Mf":E.integrate(t, algorithm="fricas")})
Sols.update({"Mg":E.integrate(t, algorithm="giac")})
Sols.update({"Mm":mathematica.Integrate(E,t).sage()}
Sols.update({"Ms":E.integrate(t, algorithm="sympy")}))

We have now six expressions (four pairwise distinct expressions) for a primitive of your expression $E$:

Sols
{'Mn': 1/2*sqrt(-t^2 + 1)*c*t + 1/2*c*arcsin(t) - 1/3*(-t^2 + 1)^(3/2),
 'Mp': 1/2*sqrt(-t^2 + 1)*c*t + 1/2*c*arcsin(t) - 1/3*(-t^2 + 1)^(3/2),
 'Mf': 1/6*(3*c*t^5 + 2*t^6 - 15*c*t^3 - 12*t^4 + 12*c*t + 12*t^2 - 6*(3*c*t^2 - (c*t^2 - 4*c)*sqrt(-t^2 + 1) - 4*c)*arctan((sqrt(-t^2 + 1) - 1)/t) + 3*(3*c*t^3 + 2*t^4 - 4*c*t - 4*t^2)*sqrt(-t^2 + 1))/(3*t^2 - (t^2 - 4)*sqrt(-t^2 + 1) - 4),
 'Mg': 1/2*c*arcsin(t) + 1/6*((3*c + 2*t)*t - 2)*sqrt(-t^2 + 1),
 'Mm': 1/2*c*arcsin(t) + 1/6*(3*c*t + 2*t^2 - 2)*sqrt(-t^2 + 1),
 'Ms': 1/2*sqrt(-t^2 + 1)*c*t + 1/3*sqrt(-t^2 + 1)*t^2 + 1/2*c*arcsin(t) - 1/3*sqrt(-t^2 + 1)}

More easily seen and understood via \LaTeX:

$$ \begin{align} Mn &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, {\left(-t^{2} + 1\right)}^{\frac{3}{2}}\\Mp &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, {\left(-t^{2} + 1\right)}^{\frac{3}{2}}\\Mf &: \frac{3 \, c t^{5} + 2 \, t^{6} - 15 \, c t^{3} - 12 \, t^{4} + 12 \, c t + 12 \, t^{2} - 6 \, {\left(3 \, c t^{2} - {\left(c t^{2} - 4 \, c\right)} \sqrt{-t^{2} + 1} - 4 \, c\right)} \arctan\left(\frac{\sqrt{-t^{2} + 1} - 1}{t}\right) + 3 \, {\left(3 \, c t^{3} + 2 \, t^{4} - 4 \, c t - 4 \, t^{2}\right)} \sqrt{-t^{2} + 1}}{6 \, {\left(3 \, t^{2} - {\left(t^{2} - 4\right)} \sqrt{-t^{2} + 1} - 4\right)}}\\Mg &: \frac{1}{2} \, c \arcsin\left(t\right) + \frac{1}{6} \, {\left({\left(3 \, c + 2 \, t\right)} t - 2\right)} \sqrt{-t^{2} + 1}\\Mm &: \frac{1}{2} \, c \arcsin\left(t\right) + \frac{1}{6} \, {\left(3 \, c t + 2 \, t^{2} - 2\right)} \sqrt{-t^{2} + 1}\\Ms &: \frac{1}{2} \, \sqrt{-t^{2} + 1} c t + \frac{1}{3} \, \sqrt{-t^{2} + 1} t^{2} + \frac{1}{2} \, c \arcsin\left(t\right) - \frac{1}{3} \, \sqrt{-t^{2} + 1} \end{align}$$

(one can ensure that maxima's solution for $c>1$ is identical to the one found for $c<-1$:

with assuming(c>1): bool(E.integrate(t)==Sols.get("Mp"))
True).

But all these expressions turn out to derivate to $E$:

[(Sols.get(u).diff(t)/E).canonicalize_radical() for u in Sols.keys()]
[1, 1, 1, 1, 1, 1]

Different integrators have different strategies for integration, leading to different primitives. Barring discontinuities, these four expressions should differ by a constant.

edit flag offensive delete link more

Comments

thank you! this is great!

Sha gravatar imageSha ( 2020-02-13 04:54:27 +0200 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2019-07-22 10:43:22 +0200

Seen: 309 times

Last updated: Jul 24 '19