# How can a series be equal to this integral?

• opus
In summary, the conversation discusses finding the series that is equal to the integral of ##\int_0^1 x^2\cos(x^3)dx##. The solution involves finding a Taylor series for the integrand and using the fact that for power series, the integral can be put inside the sum. The resulting series can be evaluated and simplified using the fact that ##6n+3 = 3(2n+1)##. The experts clarify that the center of the Taylor series is relevant when the integral has bounds other than [0,1].
opus
Gold Member

## Homework Statement

Determine the series that is equal to the integral ##\int_0^1 x^2\cos(x^3)dx##

## The Attempt at a Solution

So I didn't really know what I was doing but I did end up with the correct solution.
What I did was to find a Taylor Series for the integrand, this turned out to be ##\sum_{n=0}^\infty (-1)n\frac{x^{6n+2}}{(2n)!}##

Now I have ##\int_0^1\sum_{n=0}^\infty (-1)^n\frac{x^{6n+2}}{(2n)!}dx##
After some research, I found that for power series (for which a Taylor series is) you can put the integral inside the sum (if this is incorrect, please let me know).

Now I have ##\sum_{n=0}^\infty (-1)^n\frac{1}{(2n)!}\int_0^1 x^{6n+2}dx##
This looks like a mess, but not sure how I should have grouped this the proper way. That is, should I have put everything after ##\sum## in brackets?

Eventually, I end up evaluating and I get ##\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}##

Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.

Also, apologies for the hideous notation. I don't really know what goes where, and this type of problem seems very "choppy" for me right now. So everything is pretty far from nice to look at

Also again...
Here's an attachment of my work in case there were specific things that need to be addressed.

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opus said:
...
Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.

I don't understand why you think that a series can never equal an integral. A series in the case it converges, it is equal to a real number. A definite integral also is equal to a real number... Why can't we have that two real numbers are equal?

opus
Delta2 said:
I don't understand why you think that a series can never equal an integral. A series in the case it converges, it is equal to a real number. A definite integral also is equal to a real number... Why can't we have that two real numbers are equal?
Here’s what I mean that goes along with the lat part of my first post that is part of my confusion. Not the “bargraph” like piece represents a series.

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Ok I think I understand now what you trying to say. Still it isn't quite true for two reasons:
1) The interval of integration can be anything when we take our integral , it is not necessarily have to be from 0 to +infinite. And indeed in our case the interval of integration is from 0 to 1.
2) I can define a step-wise function such that ##f(x)=a_n , x\in[n,n+1)## and the integral of that function from 0 to +infinite ##\int_0^{+\infty}f(x)dx =\sum a_n##would be equal to the series .

Last edited:
opus
Don't overthink this.
opus said:
Determine the series that is equal to the integral ##\int_0^1 x^2\cos(x^3)dx##
Evaluate the integral (it's an easy substitution) and you get a number for which you can find a Taylor series. Voila!

opus
opus said:
Eventually, I end up evaluating and I get
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}$$
Use the fact that ##6n+3 = 3(2n+1)##. You should be able to identify the resulting series.

opus
Delta2 said:
Ok I think I understand now what you trying to say. Still it isn't quite true for two reasons:
1) The interval of integration can be anything when we take our integral , it is not necessarily have to be from 0 to +infinite. And indeed in our case the interval of integration is from 0 to 1.
2) I can define a step-wise function such that ##f(x)=a_n , x\in[n,n+1)## and the integral of that function from 0 to +infinite ##\int_0^{+\infty}f(x)dx =\sum a_n##would be equal to the series .
Ok yes that answered my question, thank you.

Delta2
Mark44 said:
Don't overthink this.

Evaluate the integral (it's an easy substitution) and you get a number for which you can find a Taylor series. Voila!
So let me ask this. Say the integral has bounds [0,2]. When finding a Taylor series, is the center of the Taylor series relevant? I want to say it definitely would be, but I don't have a good argument for this.

vela said:
Use the fact that ##6n+3 = 3(2n+1)##. You should be able to identify the resulting series.
Thanks!

The series you have is not the Riemann sum of the integral. It is just a number of different contributions to it, integration is linear.

Compare with how you would integrate 1+x ...

From what I understood, his point was that: how can a series be equal to the integral of any function from 0 to infinity, since the area under a random curve that extends from 0 to infinity, cannot equal to a series because a series is the area that is just a sum of bars and it just approximates the area under that random curve, that is it approximates our integral (in best case scenario). I believe I answered this point at post #6.

opus said:
So let me ask this. Say the integral has bounds [0,2]. When finding a Taylor series, is the center of the Taylor series relevant? I want to say it definitely would be, but I don't have a good argument for this.
I don't think the center is relevant here. For example, the Maclaurin series for ##e^x## is ##1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + \dots##, so it's straightforward to find, say, ##e^1##.

Delta2 said:
From what I understood, his point was that: how can a series be equal to the integral of any function from 0 to infinity
Not relevant here. The integral in question is over the interval [0, 1].
Delta2 said:
, since the area under a random curve that extends from 0 to infinity, cannot equal to a series because a series is the area that is just a sum of bars and it just approximates the area under that random curve, that is it approximates our integral (in best case scenario). I believe I answered this point at post #6.

opus said:

## Homework Statement

Determine the series that is equal to the integral ##\int_0^1 x^2\cos(x^3)dx##

## The Attempt at a Solution

So I didn't really know what I was doing but I did end up with the correct solution.
What I did was to find a Taylor Series for the integrand, this turned out to be ##\sum_{n=0}^\infty (-1)n\frac{x^{6n+2}}{(2n)!}##

Now I have ##\int_0^1\sum_{n=0}^\infty (-1)^n\frac{x^{6n+2}}{(2n)!}dx##
After some research, I found that for power series (for which a Taylor series is) you can put the integral inside the sum (if this is incorrect, please let me know).

Now I have ##\sum_{n=0}^\infty (-1)^n\frac{1}{(2n)!}\int_0^1 x^{6n+2}dx##
This looks like a mess, but not sure how I should have grouped this the proper way. That is, should I have put everything after ##\sum## in brackets?

Eventually, I end up evaluating and I get ##\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}##

Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.

If your problem was to evaluate the indefinite integral ##\int x^2 \cos(x^3) \, dx## then, indeed, you would get an infinite series involving ##x##. However, you have a definite integral in which ##x = 1##, so that is why you don't see an ##x## in your answer. No rocket science involved here!

Ray Vickson said:
If your problem was to evaluate the indefinite integral ##\int x^2 \cos(x^3) \, dx## then, indeed, you would get an infinite series involving ##x##.
This is true, but somewhat misleading. Finding an antiderivative is very straightforward, and results in a simple function. As an additional step, you can rewrite that function as an infinite series.
Ray Vickson said:
However, you have a definite integral in which ##x = 1##, so that is why you don't see an ##x## in your answer. No rocket science involved here!

## 1. How can a series be equal to this integral?

The equality between a series and an integral is based on the concept of convergence. If a series converges, its sum can be approximated by an integral. This is known as the integral test.

## 2. What is the integral test?

The integral test is a method used to determine whether a series converges or diverges. It states that if a series is positive, continuous, and decreasing, then the sum of the series can be approximated by the integral of the function that defines the series.

## 3. Can any series be represented by an integral?

No, not all series can be represented by an integral. The series must meet certain criteria, such as being positive, continuous, and decreasing, in order for the integral test to be applicable.

## 4. How is the integral test used in practice?

The integral test is used to determine the convergence of a series. If the integral of the function that defines the series converges, then the series also converges. If the integral diverges, then the series also diverges.

## 5. Are there other methods for determining the convergence of a series?

Yes, there are other methods such as the comparison test, ratio test, and root test. These methods compare the given series to a known series and use limit calculations to determine convergence or divergence.

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