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I'm trying to determine that f(x) = (a/2)*e^(-a|x|) is a pdf for which I have tried to calcuate the integral from -infinity to +infinity but I am no getting a result that converges(even the wolfram alpha widget said the integral doesn't converge). How do I correcly implement this?

why the the software can't solve the integral like ∫xdlnx?

Thanks in advance for your help.

hi.please help me for solve this integral


print(`output redirected...`); # input placeholder
evalf(Int(diff(Phi(x, theta, z, t), theta, theta), z = -(1/2)*h .. (1/2)*h));
print(`output redirected...`); # input placeholder
| / / /3 \\
| |(Phi[0]) . |sin|- Pi (x - 6)|| . (cos(theta))
| \ \ \2 //

cos(2 Pi z)| . (sin(omega t)) dz


How to calculate the integral

(symbolically or/and numerically) with Maple?


I would like to pay attention to an article " Sums and Integrals: The Swiss analysis knife " by Bill Casselman, where the Euler-Maclaurin formula is discussed.  It should be noted all that matter is implemented in Maple through the commands bernoulli and eulermac. For example,


eulermac(1/x, x);


eulermac(sin(x), x, 6);

BTW, as far as I know it, this boring stuff is substantially used in modern physics. The one is considered in

Ronald Graham, Donald E. Knuth, and Oren Patashnik, Concrete mathematics, Addison-Wesley, 1989.

The last chapter is concerned with the Euler-MacLaurin formula.


Hi everyone,

I have a great problem with the evaluation of following definite integral

> restart;

> int((t-x)^2)/(1+2t+(1/2)t^2-ln(t^2+2t+2)t-ln(t^2+2t+2)+arctan(1+t)t^2+2arctan(1+t)t+ln(2)t+ln(2)-(pi/4)t^2-(pi/2)t)^2,t=0..x)

I have tried different classical commands but Maple doesn't give an answer. Probably, it's just a silly fault.

Does anyone knows how to solve it?


Hi all,


I wonder if any of you guys can figure out why this integral is taking more than 2 hours to return a result??

I had similar problems in the past that were fixed creating a procedure, changing the solver and in case of trigonometric functions, changing the argument from float to rational number. 

Here it





ms := (1.141448075+9.645873109*10^(-11)*I)*(MeijerG([[], []], [[1/4, 1/4], [1/2, 0]], .3956862293*(-.70*x+1)^4)+(2.148399968-2.148399963*I)*MeijerG([[], []], [[1/4], [1/2, 1/4, 0]], -.3956862293*(-.70*x+1)^4)+(-12.48809431-3.188863063*10^(-9)*I)*MeijerG([[], []], [[0], [1/2, 1/4, 1/4]], -.3956862293*(-.70*x+1)^4)+(4.061500400*10^(-10)-8.649913391*I)*MeijerG([[], []], [[1/2], [1/4, 1/4, 0]], -.3956862293*(-.70*x+1)^4));

b := .7;



"H3p:=proc(ms,b) local Hcub;  Hcub := evalf(Int(diff((diff(ms, x))*(int((-x*b+1)*(int((ms')^2, x = 0 .. x)), x = 1 .. x)), x)*ms, x=0..1,method = _d01ajc)): return Hcub; end proc:   "



st := time(); ``*H3p(ms, b); time()-st




Hi, I'm currently from Mathematica and it starts to disappouint me because it cannot do what I can.

Can somebody try to calculate undefinite integral in Maple for x variable. a and b are parameters.


If Maple can do that then I would switch to Maple comunity.






Does not work.





But with some effect,



Could this possibly be improved in the next release?


I want to solve the int((y4+by2+c)-1/2,y)-x and find y=h(x), where b and c are constants s.t. c>b2/4. Maple gives me complex Jacobi elliptic function as a result. But I am not sure that this integral has complex value. Am I doing something wrong or the result is really a complex valued function? Thanks.

Indeed my main question is: Plot y=y(u) where we have these two relations: int((y4+by2+c)-1/2,y)=x and find y=h(x). Then evaluate int((h(x)-B)-1/2,x)=u and find x=g(u). By using these relations plot y=y(u). :)

Here B is an arbitrary constant, but if necessary we can define a value for it. All the variables and constants are real.

I hope I manage to express myself. Thanks again.


Int(piecewise(t < T1, exp((1/2)*t*(1+2*I-I*sqrt(3))), t < T2, -1000*exp((1/2)*t*(1+2*I-I*sqrt(3)))*(-1/1000+T1-t), T2 <= t, -1000*exp((1/2)*t*(1+2*I-I*sqrt(3)))*(-1/1000-T2+T1)), t)



Let us consider the definite integral

J:=int(abs(x-(-x^5+1)^(1/5)), x = 0 .. 1);

Maple fails with it, Mathematica 10.1 finds it in terms of  special functions. Let us look at the integrand:
plot(x-(-x^5+1)^(1/5), x = 0 .. 1);

We see the expression under the modulus changes its sign at the unique point of RealRange(0,1). Therefore



J:= int(-x+(-x^5+1)^(1/5), x = 0 .. (1/2)*2^(4/5))+int(x-(-x^5+1)^(1/5), x = (1/2)*2^(4/5) .. 1);

which outputs a complicated expression

(1/8)*2^(4/5)*(4*hypergeom([-1/5, 1/5], [6/5], 1/2)-2^(4/5))+(1/2)*2^(4/5)*((1/2)*2^(1/5)-(1/4)*2^(4/5))-(1/25)*Pi*csc((1/5)*Pi)*(-(25/2)*sin((1/5)*Pi)*GAMMA(4/5)*2^(4/5)*hypergeom([-1/5, 1/5], [6/5], 1/2)/Pi+(5/4)*sec((3/10)*Pi)*cos((1/10)*Pi)*2^(3/5)*Pi^(1/2)*csc((3/10)*Pi)/GAMMA(7/10))/GAMMA(4/5).

At the same time we have

int(abs(x-(-x^5+1)^(1/5)), x = 0 .. 1, numeric);


How to obtain 1/2 symbolically?

according to help on timelimit

"Note: For efficiency reasons, the timelimit bound is ignored while in built-in routines."

Which is not very useful, since I want to limit  int() to some CPU time.

There are some integrals that can hang Maple easily. I'd like to set some CPU time on an int() and
have it terminate with error, but I am not able to find how to do that.

For example this


Will hangs Maple.
Is there a trick some expert here can show to limit the CPU time on a build in operation?
May be some package or other command can do this?

I am only interested in int() now, but if it can work also on dsolve, that will be good.

thank you
ps. Mathematica supports putting time constraint on build-in commands. So I do
not see why Maple can't also do the same.

of the cut-off sphere



Of course, with Maple.

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