Respuesta :
Answer:
- 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
The integration of the arctan (x)/x is C + Σ [(-1)^n x^(2n+1)]/(2n+1)² where n is from 0 to ∞. The radius of convergence is R = 1/x
What is integration?
It is the reverse of differentiation.
The indefinite integral is a power series that will be
[tex]\rm \int \dfrac{\tan^{-1}x }{ x} \ dx[/tex]
We know that the arctan is given as
[tex]\rm \tan^{-1} x = \dfrac{\sum [(-1)^n \ x^{2n+1}]}{(2n+1)}\\\\\\\\dfrac{\tan^{-1} x }{x} = \dfrac{\sum [(-1)^n \ x^{2n})]}{(2n+1)}\\\\\\\dfrac{\tan^{-1} x }{x} = \dfrac{\sum [(-1)^n \ ]}{(2n+1)} \ \ x^{2n}[/tex]
Then we have
[tex]\rm \int \dfrac{\tan^{-1} x }{x}\ dx= \int \dfrac{\sum [(-1)^n \ ]}{(2n+1)} \ \ x^{2n} dx\\\\\\\int \dfrac{\tan^{-1} x }{x} \ dx = \dfrac{\sum [(-1)^n \ ]}{(2n+1)} \ \int x^{2n}\\\\\\\int \dfrac{\tan^{-1} x }{x} \ dx = \int \dfrac{\sum [(-1)^n \ ]}{(2n+1)^2} x^{2n + 1} + C[/tex]
To obtain the radius of convergence, then the ration test
[tex]\rm R = \displaystyle \lim_{n \to \infty} \dfrac{a_n}{a_{n+1}}\\\\\\a_n = \dfrac{(-1)^n \ x^{2n+1} }{(2n+1)^2}\\\\\\a_{n+1} = \dfrac{(-1)^{n+1} \ x^{2(n+1)+1} }{(2(n+1)+1)^2}\\\\\\ \dfrac{a_n}{a_{n+1}} = \dfrac{1}{x}(1+\dfrac{2}{2n+1})[/tex]
Then we have
[tex]\rm R = \displaystyle \lim_{n \to \infty} \dfrac{1}{x}(1+\dfrac{2}{2n+1})\\\\\\R = \dfrac{1}{x}[/tex]
More about the integration link is given below.
https://brainly.com/question/18651211
