## 56 Reputation

17 years, 26 days

## Unable to plot an integral function cont...

Maple

Here are my codes:

restart;
Delta := proc (beta) options operator, arrow; sqrt(1-1/beta^2) end proc;
T := proc (`&epsilon;`, Z) options operator, arrow; piecewise(`&epsilon;` < Delta(beta) and Delta(beta) < `&epsilon;`, 2*Delta(beta)^2/(`&epsilon;`^2+(Delta(beta)^2-`&epsilon;`^2)*(2*Z^2+1)^2), `&epsilon;` < -Delta(beta) or Delta(beta) < `&epsilon;`, 2*abs(`&epsilon;`)/(abs(`&epsilon;`)+sqrt(`&epsilon;`^2-Delta(beta)^2)*(2*Z^2+1))) end proc;

## A plot fail problem...

Maple 15

Hi guys, my Maple version is 15.01, when I want to do the following plots, the Maple pops up this failure "Kernal Connenction Lost":

Here're the corresponding input commands:

J := u -> evalf(Int(1/(sqrt(x^2+u^2)*(exp(sqrt(x^2+u^2))+1)), x = 0 .. infinity));

plot(exp(-2*J(Delta/(.4))), Delta = 0 .. 1);

Does anyone encouter the same problem if you try the commands? I wonder how to fix it, thank you.

## How do I get this answer?...

Maple

It is very stranger that these two statements are not equative:

evalf(Int(x*sin(x)/(1+(cos(x))^4),x=-Pi/2..Pi/2));

1.845096351

evalf(int(x*sin(x)/(1+(cos(x))^4),x=-Pi/2..Pi/2));

1.323830675+0.*I

I think the first one is correct. In fact, the exact result is

2*hypergeom([1/4, 1/2, 1, 1], [3/4, 5/4, 5/4], -1);

u can evalf the above answer to check it. In fact, Maple cannot get this analytic answer, how do I get it?

## Time spend on calculating...

Why does Maple take (much) more time on calculating than Mathematica? For example, some time ago, I calculate this integration: int((sin(x)/x)^100,x=-infinity..infinity); although Maple is able to give the correct answer, but it takes about 5 minutes to achieve this answer. However, when I did the same thing in Mathematica, the time spent on calculating is less than one minute. Why is there such a notable distinction between the two computer algebra systems? PS: Versions of the two softwares that I use are Maple 11 and Mathematica 6.

## How to get this answer?...

Maple
Consider a sum: Sum((-1)^(i+1)*Sum(1/j,j=1..i)*x^i,i=1..infinity); value(%); The result given by Maple 11 is: sum((-1)^(i+1)*(Psi(i+1)+gamma)*x^i, i = 1 .. infinity) However, from an calculus book I know the answer is: log(1+x)/(1+x). How can I get this result from Maple? Whether should I give some assumptions? Best wishes.
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