int_0^infty cos (pi x)dx=
Substitute u=pi x => du=pi dx => dx=(du)/pi, u_l=pi cdot 0=0, u_u=pi cdot infty=infty. (u_l and u_u are lower and upper bound respectively).
1/pi int_0^infty cos u du=1/pi sin u|_0^infty=1/pi(lim_(u to infty)sin u-sin0)
The integral does not converge (it diverges) because lim_(u to infty) sin u does not exist.
The image below shows the graph of the function (blue) and area between it and x-axis representing the value of integral (green positive and red negative). We can see that any such integral (with infinite bound(s)) of periodic function will diverge.
Tuesday, October 15, 2013
int_0^oo cos(pix) dx Determine whether the integral diverges or converges. Evaluate the integral if it converges.
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