Evaluate the indefinite integral as a power series.

∫ tan−1(x) / x dx

[infinity]

f(x) = C + Σ

n = 0

What is the radius of convergence R?

- The integral of arctan(x)/x =

C + Σ [(-1)^n.x^(2n+1)]/(2n+1)². (From n = 0 to infinity).

- The radius of convergence is R = 1/x

Step-by-step explanation:

First note that

tan^(-1)x = arctan(x)

And

arctan(x) = Σ [(-1)^n. x^(2n+1)]/(2n+1). From n = 0 to infinity

arctan(x)/x = Σ [(-1)^n. x^(2n)]/(2n+1). From n = 0 to infinity

∫arctan(x)/x dx = ∫{Σ [(-1)^n. x^(2n)]/(2n+1). From n = 0 to infinity}dx

= ∫{Σ [(-1)^n]/(2n+1) .From n = 0 to infinity}.∫x^(2n)dx

= {Σ [(-1)^n]/(2n+1) .From n = 0 to infinity}.x^(2n+1)/(2n+1) + C

= C + Σ [(-1)^n.x^(2n+1)]/(2n+1)². (From n = 0 to infinity).

To obtain the radius of convergence, we apply the ration test

R = Limit as n approaches infinity |a_n/a_(n+1)|

a_n = (-1)^n.x^(2n+1)]/(2n+1)²

a_(n+1) = (-1)^(n+1).x^(2(n+1)+1)]/(2(n+1)+1)²

|a_n/a_(n+1)| = (2(n+1) + 1)²/(2n+1)².x

= (1/x)[1 + 2/(2n+1)]

R = Limit as n approaches infinity |a_n/a_(n+1)|

= R = Limit as n approaches infinity (1/x)[1 + 2/(2n+1)]

R = 1/x

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

step-by-step explanation:

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