# How do you find a power series converging to f(x)=int ln(1+t^2) dt from [0,x] and determine the radius of convergence?

Jul 25, 2017

Substitute the series for $\ln \left(1 + {t}^{2}\right)$ and integrate term-by-term to get ${x}^{3} / 3 - {x}^{5} / 10 + {x}^{7} / 21 - {x}^{9} / 36 + \cdots$. The radius of convergence is 1.

#### Explanation:

The Taylor series for $\ln \left(1 + x\right)$ centered at $x = 0$ can be found by integrating the series for $\frac{1}{1 + x} = 1 - x + {x}^{2} - {x}^{3} + {x}^{4} - {x}^{5} + \cdots$ term-by-term and using the fact that $\ln \left(1\right) = 0$ to get $\ln \left(1 + x\right) = x - {x}^{2} / 2 + {x}^{3} / 3 - {x}^{4} / 4 + {x}^{5} / 5 - {x}^{6} / 6 + \cdots$.

Next, replace $x$ with ${t}^{2}$ to get $\ln \left(1 + {t}^{2}\right) = {t}^{2} - {t}^{4} / 2 + {t}^{6} / 3 - {t}^{8} / 4 + {t}^{10} / 5 - {t}^{12} / 6 + \cdots$.

Now integrate this term-by-term, from $t = 0$ to $t = x$ to get

$\setminus {\int}_{0}^{x} \ln \left(1 + {t}^{2}\right) \mathrm{dt} = {x}^{3} / 3 - {x}^{5} / 10 + {x}^{7} / 21 - {x}^{9} / 36 + \cdots$

The original radius of convergence for $\frac{1}{1 + x}$ was 1, and that does not change upon substitution of ${t}^{2}$ for $x$ or upon either integration.

The interval of convergence does get "expanded" by one point, however. The interval of convergence for the final answer is $\left(- 1 , 1\right]$ (include 1 but do not include $- 1$).

Jul 25, 2017

${\int}_{0}^{x} \ln \left(1 + {t}^{2}\right) \mathrm{dt} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 3} / \left(\left(n + 1\right) \left(2 n + 3\right)\right)$

for $\left\mid x \right\mid < 1$

#### Explanation:

Note that:

$\frac{d}{\mathrm{dx}} \left(\ln \left(1 + {x}^{2}\right)\right) = \frac{2 x}{1 + {x}^{2}}$

Now:

$\frac{1}{1 + {x}^{2}}$

can be expressed as the sum of a geometric series of ratio $\left(- {x}^{2}\right)$ with ${x}^{2} < 1$:

$\frac{1}{1 + {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- {x}^{2}\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n}$

and then:

$\frac{2 x}{1 + {x}^{2}} = 2 x {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n} = 2 {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$ for $\left\mid x \right\mid < 1$

Inside the interval of convergence we can integrate term by term to have a power series with the same radius of convergence:

$\ln \left(1 + {x}^{2}\right) = {\int}_{0}^{x} \frac{2 t}{1 + {t}^{2}} \mathrm{dt} = 2 {\sum}_{n = 0}^{\infty} {\int}_{0}^{x} {\left(- 1\right)}^{n} {t}^{2 n + 1} \mathrm{dt}$

$\ln \left(1 + {x}^{2}\right) = 2 {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2} / \left(2 n + 2\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2} / \left(n + 1\right)$

and integrating again:

${\int}_{0}^{x} \ln \left(1 + {t}^{2}\right) \mathrm{dt} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) {\int}_{0}^{x} {t}^{2 n + 2} \mathrm{dt} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 3} / \left(\left(n + 1\right) \left(2 n + 3\right)\right)$