# Maclaurin Series for an Integral

• Frillth
In summary, the conversation discusses finding the first 3 non-zero terms of the Maclaurin series of f(x) = Integral sin(t)/t dt evaluated from 0 to x and determining if there is a simpler way to represent the integral. It is concluded that the integral is well-defined and the given form is adequate for solving the problem.
Frillth
For my homework in my class, I'm supposed to find the first 3 non-zero terms of the Maclaurin series of f(x) = Integral sin(t)/t dt evaluated from 0 to x. I'm fairly sure that sint/t dt has no integral, so I'm lost in my search for the solution. Can you guys point me in the right direction?

$$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!}x^{2n+1}$$$$\frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!}x^{2n}$$

$$\int_{0}^{x} \frac{\sin x}{x}\; dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!(2n+1)}x^{2n+1}$$

Last edited:
I'm fairly sure that sint/t dt has no integral
Why do you think that? You're integrating a continuous, bounded function over a finite interval -- that's like the poster child for being integrable!

The integral most surely exists. The function

$$g(x) := \int_0^x \frac{\sin t}{t} \, dt$$

is perfectly well-defined. It just might be that there isn't a "simpler" way to represent g(x). But it doesn't matter; this form is more than adequate for solving the problem at hand.

Last edited:
I'm sorry, I didn't make myself clear. What I meant when I said that sint/t dt has no derivative is that the indefinite integral sint/t has no simple function to represent it. I looked up the integral, and wolfram just calls it sine integral.

You do see that this form is adequate for the problem, though, right? If not... then where do you see an obstacle?

## What is a Maclaurin Series for an Integral?

A Maclaurin Series for an Integral is a mathematical representation of a function as an infinite sum of terms, derived from a specific point (usually x=0) and its derivatives. It is used to approximate the value of a function at any point within its interval of convergence.

## How is a Maclaurin Series for an Integral calculated?

A Maclaurin Series for an Integral is calculated by first finding the indefinite integral of the given function. Then, the resulting integral is evaluated at the specific point (x=0) to obtain the value of the constant term. Finally, the higher-order derivatives of the integral are evaluated at x=0 and used to construct the remaining terms in the series.

## What is the interval of convergence for a Maclaurin Series for an Integral?

The interval of convergence for a Maclaurin Series for an Integral is the set of all values of x for which the series converges. In most cases, this interval is a symmetric interval centered at the point of expansion (x=0). However, it is important to check the endpoints of the interval for convergence as well.

## What are the major applications of Maclaurin Series for an Integral?

Maclaurin Series for an Integral have many applications in mathematics and physics. They are used to approximate the value of a function, as well as to simplify complicated expressions and solve differential equations. They are also used in the fields of engineering, economics, and statistics, among others.

## What are some common mistakes to avoid when working with Maclaurin Series for an Integral?

One common mistake is to forget to check the interval of convergence and assume that the series will always converge. It is also important to carefully evaluate the higher-order derivatives at the point of expansion (x=0) to ensure accurate terms in the series. Additionally, it is important to remember that a Maclaurin Series is only an approximation and may not accurately represent the function for all values of x.

• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
23
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
985
• Calculus and Beyond Homework Help
Replies
1
Views
922
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
389
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
16
Views
2K