From this expression I want to extract only terms with the variable x.

F := (1/2/Pi+Q*cos(k*x))^2*(1/2/Pi+Q*cos(k*y))/(1+beta^2*(1/2/Pi+Q*cos(k*x))^2);

I don't want to manually write it out rather extract the terms programmaticallly. I want to be left with

(1/2/Pi+Q*cos(k*x))^2 / (1+beta^2*(1/2/Pi+Q*cos(k*x))^2)

I have tried the collect fucntion but that did not seem to do it. Any suggestions...

Hi guys,

I would be ever so grateful if someone could asssist me with the following matter.

I am trying to integrate the following expression but maple does not seem to be able to do it directly

Int((1/(2*Pi) + Q* cos(k*phi))^2/ (1+ beta^2*((1/(2*Pi)) + Q* cos(k*phi) )^2),phi=-Pi..Pi);

So I convert the expression to an exponential form as follows and evaluate that

 numerator:=(1/2/Pi+Q*(1/2*exp(I*k*phi...

Please help me with the following.

I have this function

y(theta,t):=1/4*Pi + A*sin(m*theta) + A*sin(m*theta)* (exp(-1/4 *alpha^2*beta^2)-2)/(4*Pi)*t;

Hi guys please help me with the following.

When I evaluate the integral below

int(1/8*2^(1/2)*exp(-1/2*x^2/sigma^2)/(Pi^(5/2)*sigma)+1/8*2^(1/2)*exp(-1/2*x^2/sigma^2)*Q*exp(I*k*x)*exp(I*k*theta)/(Pi^(3/2)*sigma)+1/8*2^(1/2)*exp(-1/2*x^2/sigma^2)*Q/(Pi^(3/2)*sigma*exp(I*k*x)*exp(I*k*theta)), x = (-infinity .. infinity));    --------(1)

I get an expression with things like 

csgn(conjugate(sigma)) and 

Consider the expression below.

r:=1/2 * exp(I*k*theta - 1/2*k^2*sigma^2)* sigma*sqrt(2*Pi) + 1/2 * exp(-I*k*theta - 1/2*k^2*sigma^2)* sigma*sqrt(2*Pi);       -------(1)

BY HAND I can factor out the sigma square root of 2 pi and one of the exponents to give.

sigma * sqrt(2*Pi...