How can a series be equal to this integral?

opus

Gold Member
1. The problem statement, all variables and given/known data
Determine the series that is equal to the integral $\int_0^1 x^2\cos(x^3)dx$

2. Relevant equations

3. 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.

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opus

Gold Member
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 opus

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

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• 164 KB Views: 58

Delta2

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Gold Member
...
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 cant we have that two real numbers are equal???

• opus

opus

Gold Member
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 cant 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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Delta2

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Gold Member
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

Mark44

Mentor
Don't overthink this.
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

vela

Staff Emeritus
Homework Helper
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

opus

Gold Member
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

opus

Gold Member
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.

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

Orodruin

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2018 Award
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 ...

Delta2

Homework Helper
Gold Member
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.

Mark44

Mentor
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, $e^1$.

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.

Ray Vickson

Homework Helper
Dearly Missed
1. The problem statement, all variables and given/known data
Determine the series that is equal to the integral $\int_0^1 x^2\cos(x^3)dx$

2. Relevant equations

3. 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!

Mark44

Mentor
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!

"How can a series be equal to this integral?"

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