# Evaluate the indefinite integral as a power series. f(t)=\int \frac {\ln(11-t)}{4t}dt. What is...

## Question:

Evaluate the indefinite integral as a power series {eq}f(t)=\int \frac {\ln(11-t)}{4t}dt. {/eq} What is the radius of convergence {eq}R {/eq}?

## Indefinite Integral:

There are various indefinite integral whose integrand is not solved using the normal set of standard formulas of the integral method, but to solve those we need to apply the concept of the Taylor series expansion formula.

So the indefinite integral given here is:

{eq}f(t)=\int \frac{\ln(11-t)}{4t}dt {/eq}

Now we will first find the power series for the integrand:

{eq}\frac{\ln(11-t)}{4t} {/eq}

by the Taylor series formula:

{eq}\ln \left(1+x\right)= \sum _{n=1}^{\infty \:}\left(-1\right)^{n+1}\frac{x^n}{n}\\ {/eq}

So we will have:

{eq}\Rightarrow \ln \left(11-t\right)\\ =\ln 11\left(1-\frac{t}{11}\right)\\ =\ln 11 + \ln \left(1-\frac{t}{11}\right)\\ =\ln 11 + \sum _{n=1}^{\infty \:}\left(-1\right)^{n+1}\frac{(-\frac{t}{11})^n}{n}\\ {/eq}

So now the integral of this series is:

{eq}\int \frac{\ln(11-t)}{4t}dt\\ =\int \ln 11 dt+\int \sum _{n=1}^{\infty \:}\left(-1\right)^{n+1}\frac{(-\frac{t}{11})^n}{n} dt\\ =t \ln 11 - 11 \sum _{n=1}^{\infty \:}\left(-1\right)^{n+1}\frac{(-\frac{t}{11})^{n+1}}{n(n+1)}+C\\ {/eq}

Now, this is the required integral.

Also, we have this as the converging integral is:

{eq}|\frac{t}{11}| \le 1\\ \Rightarrow -11\le \:t\le \:11 {/eq}

So we have the radius of convergence as:

{eq}R=11 {/eq}