The first thing to do is get rid of the arctan. We can do this by realizing that as x -> oox→∞, arctan(x)->pi/2arctan(x)→π2. We can see this from the graph (or just by knowing that tan(x) has horizontal asymptotes at x=pi/2x=π2):
This means that arctan(x)arctan(x) on [0,oo) <= pi/2[0,∞)≤π2 and therefore
int_0^oo arctan(x)/(2+e^x)dx<= pi/2 int_0^oo 1/(2+e^x)dx∫∞0arctan(x)2+exdx≤π2∫∞012+exdx
This is still a bit tricky to integrate, so we can simplify further:
2+e^x > e^x2+ex>ex which means
1/(2+e^x)<1/e^x12+ex<1ex so,
pi/2 int_0^oo 1/(2+e^x)dx <= pi/2 int_0^oo 1/(e^x)dxπ2∫∞012+exdx≤π2∫∞01exdx
Now we have:
int_0^oo arctan(x)/(2+e^x)dx <= pi/2 int_0^oo 1/(e^x)dx∫∞0arctan(x)2+exdx≤π2∫∞01exdx
and pi/2 int_0^oo 1/(e^x)dxπ2∫∞01exdx is something we can integrate!
pi/2 int_0^oo 1/(e^x)dx = lim_(t->oo) pi/2 int_0^t e^-x dx
=lim_(t->oo) pi/2[-e^-x]_0^t
= lim_(t->oo) pi/2[-e^-t-(-e^0)]
= pi/2[0+1]
=pi/2 so
int_0^oo arctan(x)/(2+e^x)dx <= pi/2
and therefore it is convergent :)
