Integrate the following using infinite $\bb\text(series)$ ?

Question

(a) $\int(\arctan(4x))/(x^-7)dx$

1764 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-10T23:31:45

$int arctan(4x)/x^(-7)dx = sum_(n=0)^oo (-1)^n 2^(4n+2)/((2n+1)(2n+9) )x^(2n+9) +C$

Start from:

$arctan(4x) = int_0^(4x) dt/(1+t^2)$

Expand now the integrand using the geometric series:

$arctan(4x) = int_0^(4x) sum_(n=0)^oo (-1)^nt^(2n)dt$

for $abs (4x) < 1$ the series is absolutely convergent and we can integrate term by term:

$arctan(4x) = sum_(n=0)^oo (-1)^nint_0^(4x) t^(2n)dt$

$arctan(4x) = sum_(n=0)^oo (-1)^n (4x)^(2n+1)/(2n+1)$

$arctan(4x) = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)x^(2n+1)$

Dividing by $x^(-7)$ means multiplying by $x^7$ and we can do it again term by term:

$arctan(4x)/x^(-7) = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)x^(2n+8)$

and integrating again term by term:

$int arctan(4x)/x^(-7)dx = sum_(n=0)^oo (-1)^n 2^(4n+2)/(2n+1)int x^(2n+8) dx$

$int arctan(4x)/x^(-7)dx = sum_(n=0)^oo (-1)^n 2^(4n+2)/((2n+1)(2n+9) )x^(2n+9) +C$