`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...

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`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.