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

4 Answers
Jul 6, 2016

int_0^{oo} (x*arctanx) dx / (1 + x^2)^2 = pi/8

Explanation:

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

Jul 6, 2016

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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Jul 6, 2016

pi/8

Explanation:

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

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Jul 6, 2016

= pi/8

Explanation:

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