How do you evaluate #int (x*arctanx) dx / (1 + x^2)^2# from 0 to infinity?

Question

Samantha Adkison · Accepted Answer

#pi/8#

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#

Aurora Alvarado · Answer

#= 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 #

Charles Anctil · Answer

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.# If this is helpful to you, then like me, have fun with math!

Samantha Adkison · Answer

#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#