How do you evaluate #int (x*arctanx) dx / (1 + x^2)^2# from 0 to infinity?
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#int_0^{oo} (x*arctanx) dx / (1 + x^2)^2 = pi/8#
#d/(dx)(arctan(x)/(1+x^2)) =1/(1 + x^2)^2 - (2 x arcTan(x))/(1 + x^2)^2 #
so
#int (x*arctanx) dx / (1 + x^2)^2 = 1/2int dx/(1 + x^2)^2- 1/2arctan(x)/(1+x^2) #
but
#1/(1 + x^2)^2 = 1/2 d/(dx)(x/(1 + x^2))+1/2 d/(dx) (arctan(x))#
concluding
#int (x*arctanx) dx / (1 + x^2)^2=(x + (x^2-1) arcTan(x))/(4 (1 + x^2)) #
but
#lim_{x->0}(x + (x^2-1) arcTan(x))/(4 (1 + x^2)) = 0#
and
#lim_{x->oo}(x + (x^2-1) arcTan(x))/(4 (1 + x^2)) = (pi/2)/4 = pi/8#
finally
#int_0^{oo} (x*arctanx) dx / (1 + x^2)^2 = pi/8#
Let #I=int_0^ oo(x*arctanx)/(1+x^2)^2dx.#
We take sbstn., #arctanx=t#, so that, #x=tant#, and, #dx=sec^2t.#
Also, #x=0 rArr t=0,# and #x rarr oo, trarr pi/2.#
Because of all these changes, now, #I# becomes,
#I=int_0^(pi/2)(t*tant*sec^2t)/sec^4tdt=int_0^(pi/2)(t*tant)/sec^2tdt#
#=int_0^(pi/2)(t)(sint/cost)cos^2tdt=int_0^(pi/2)tsintcostdt#
#=(1/2)int_0^(pi/2)tsin2tdt.................(i)#
#=1/2[t*{-cos(2t)/2}]_0^(pi/2)-(1/2)int_0^(pi/2)[1*{-cos(2t)/2}]dt#
#=-1/4[{pi/2*cospi}-0]+1/4[sin(2t)/2]_0^(pi/2)#
#=pi/8+1/8[sinpi-sin0]#
#:. I=pi/8#.
To evaluate #I# further from #(i)#, we have used the Rule of Integration by Parts for Definite Integral :
#int_a^bu*vdx=[u*intvdx]_a^b-int_a^b{(du)/dx*intvdx}dx,# with #u=t, &, v=sin2t.#
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Use the substitution #x = tan u and 1+tan^2u=sec^2 u#..
The limits become # pi/2 and oo, and dx = sec^2 u# #du#.
The given integral reduces to the form
#int u sin u cos u du=1/2intu sin 2u du#
#=(-1/4)intud(cos 2u)#
#=-1/4[u cos 2u-int cos 2u du] #, between the limits #0 and pi/2#
=. #(-1/4)(pi/2)(-1)+1/4[1/2sin 2u]#, between the limits
#=pi/8+0=pi/8#
-
#int_0^oo dx qquad (x*arctanx) / (1 + x^2)^2#
using IBP
#u = arctan x, u' = 1/(1+ x^2)#
#v' = x / (1 + x^2)^2, v = -1 /(2 (1 + x^2))#
#arctan x (-1 /(2 (1 + x^2)) ) - int_0^oo dx qquad 1/(1+ x^2)* (-1 /(2 (1 + x^2)))#
# = -arctan x /(2 (1 + x^2)) + 1/2 color{red}{int_0^oo dx qquad 1/( (1 + x^2)^2)} qquad star#
for the red bit we use #x = tan psi, dx = sec^2 psi d psi#
#implies int dpsi qquad sec^2 psi / sec^4 psi#
#= int dpsi qquad cos^2 psi#
#= 1/2 int dpsi qquad cos2 psi +1#
#= 1/2 ( 1/2 sin 2 psi + psi) #
#= 1/2 ( sin psi cos psi + psi)#
#= 1/2 ( x/sqrt{1 + x^2} * 1/sqrt{1 + x^2} + arctan x)#
#= 1/2 ( x/(1 + x^2) + arctan x)#
subbing back into #star#
# = [ -arctan x /(2 (1 + x^2)) + 1/2 (1/2 ( x/(1 + x^2) + arctan x)) ]_0^oo #
# = [ -arctan x /(2 (1 + x^2)) + 1/4 * x/(1 + x^2) + 1/4 arctan x ]_0^oo #
#= pi/8 #
