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LE_EQ.mwWhat is problem with this programme,why it does'nt calculate the values but only shows the solution with integral sign instead of calculating it, there is also arising a problem in plots

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?

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