int_0^infty sin(x/2)dx=
Substitute u=x/2 => du=dx/2 => dx=2du, u_l=0/2=0, u_l=infty/2=infty (u_l and u_u are lower and upper bound respectively).
2int_0^infty sin u du=-2cos u|_0^infty=-2(lim_(u to infty)cos u-cos 0)
The integral does not converge (it diverges) because lim_(u to infty)cos u does not exist.
The image below shows the graph of the function (blue) and area between it and -axis representing the value of the integral (green positive and red negative). We can see that any such integral (with infinite bound(s)) of periodic function will diverge.
Wednesday, August 20, 2014
int_0^oo sin(x/2) dx Determine whether the integral diverges or converges. Evaluate the integral if it converges.
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