# Taylor series of an integral function

## Homework Statement

$$\displaystyle f\left( x \right)=\int\limits_{0}^{x}{\frac{\sin t}{t}dt}$$

Calculate the Maclaurin series of third order.

## The Attempt at a Solution

What I do is:

$$\displaystyle f'\left( x \right)=\frac{\sin x}{x}$$

$$\displaystyle f''\left( x \right)=\frac{x\cos x-\sin x}{{{x}^{2}}}$$

and so on...

but if I try to evaluate theese two on x=0 it doesn't make sense. I've seen I have to transform the integrand to the taylor polynomial but why is my "method" wrong?

Thanks

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Chestermiller
Mentor

## Homework Statement

$$\displaystyle f\left( x \right)=\int\limits_{0}^{x}{\frac{\sin t}{t}dt}$$

Calculate the Maclaurin series of third order.

## The Attempt at a Solution

What I do is:

$$\displaystyle f'\left( x \right)=\frac{\sin x}{x}$$

$$\displaystyle f''\left( x \right)=\frac{x\cos x-\sin x}{{{x}^{2}}}$$

and so on...

but if I try to evaluate theese two on x=0 it doesn't make sense. I've seen I have to transform the integrand to the taylor polynomial but why is my "method" wrong?

Thanks
f' and f'' both have specific limiting values at x = 0. For example, in the limit of x approaches zero, f' = 1.

Dick
Homework Helper
It's a lot easier if you expand sin(t) in a power series. Then divide by t. You are simply trying to do it the hard way.

Zondrina
Homework Helper

## Homework Statement

$$\displaystyle f\left( x \right)=\int\limits_{0}^{x}{\frac{\sin t}{t}dt}$$

Calculate the Maclaurin series of third order.

## The Attempt at a Solution

What I do is:

$$\displaystyle f'\left( x \right)=\frac{\sin x}{x}$$

$$\displaystyle f''\left( x \right)=\frac{x\cos x-\sin x}{{{x}^{2}}}$$

and so on...

but if I try to evaluate theese two on x=0 it doesn't make sense. I've seen I have to transform the integrand to the taylor polynomial but why is my "method" wrong?

Thanks
The beauty of a Taylor series is that it allows you to express functions which are complicated to integrate in terms of an infinite sum of polynomials which are easy to integrate.

The Maclaurin series is simply the Taylor series centered at 0.

Recall that the Maclaurin series for $sinx = \sum_{n=0}^{∞}\frac{x^(2n+1)}{(2n+1)!}(-1)^n$

Using this fact you should be able to solve your integral.

Chestermiller
Mentor
It's a lot easier if you expand sin(t) in a power series. Then divide by t. You are simply trying to do it the hard way.
I agree. That's how I would have done the problem if it had been posed to me. But I wanted to show how it could also have been done using the more difficult way that the OP had pursued.

Thank you guys, now I understand ;)