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9 June, 14:01

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?

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Answers (1)
  1. 9 June, 14:29
    0
    - 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
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