Integral Evaluation: Solve $\int_{-\pi}^{\pi}x^{2014}\sin x {\rm{d}}x$

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In summary: This means that your integral is simply ##0##. In summary, the integral of ##x^{2014}## times ##\sin x## over the interval ##[-\pi, \pi]## is equal to ##0##. This is because the product of an even function and an odd function is odd, and the integral of an odd function on a symmetric interval is always equal to ##0##.
  • #1
nuuskur
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Homework Statement


[tex]\int_{-\pi}^{\pi}x^{2014}\sin x {\rm{d}}x[/tex]

Homework Equations

The Attempt at a Solution


For such problems there are probably some extremely clever solutions, but I can't see any easy way here.
If I were to find the antiderivative of this: I would eventually come up with a sum: [tex]-\frac{2014!}{0!}x^{0} \sin x + \frac{2014!}{1!}x\cos x + \frac{2014!}{2!}x^2 \sin x -\frac{2014!}{3!}x^3 \cos x + ... +\\ \frac{2014!}{2013!}x^{2013} \sin x -\frac{2014!}{2014!}x^{2014} \cos x[/tex]
[is there a way to write this sum in sigma notation?]
Are there 2015 summands?
All summands with sine in them sum up to 0, because [itex]\sin\pm\pi = 0[/itex]. and [itex]\cos\pm\pi = -1[/itex].
I have some series left from evaluating at pi and I have exactly the same series evaluated at -pi. Is the integral equal to 0? EDIT: NO, I forgot the x terms.

Is there any, more elegant solution to this problem, though?
 
Last edited:
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  • #2
An integral of the form ##\int x^n sin(x) \space dx## would usually be handled with multiple applications of integration by parts, but when ##n## is large this is obviously out of the question without the aid of a computer.

There is actually a standard form for such an integral, which relies on the notion of summing up the ##sin(x)## and ##cos(x)## terms (you have already noticed this). The standard form in terms of the gamma function is:

$$\int x^n sin(x) \space dx = - \frac{i^n}{2} \left[ \Gamma(n+1, -ix) - (-1)^n \Gamma(n+1, -ix) \right]$$

Substituting your limits should help you confirm.

EDIT: Alternatively, express ##sin(x)## in terms of its power series, bring the ##x^{2014}## in and presto I believe.
 
Last edited:
  • #3
nuuskur said:

Homework Statement


[tex]\int_{-\pi}^{\pi}x^{2014}\sin x {\rm{d}}x[/tex]

Homework Equations

The Attempt at a Solution


For such problems there are probably some extremely clever solutions, but I can't see any easy way here.
If I were to find the antiderivative of this: I would eventually come up with a sum: [tex]-\frac{2014!}{0!}x^{0} \sin x + \frac{2014!}{1!}x\cos x + \frac{2014!}{2!}x^2 \sin x -\frac{2014!}{3!}x^3 \cos x + ... +\\ \frac{2014!}{2013!}x^{2013} \sin x -\frac{2014!}{2014!}x^{2014} \cos x[/tex]
[is there a way to write this sum in sigma notation?]
Are there 2015 summands?
All summands with sine in them sum up to 0, because [itex]\sin\pm\pi = 0[/itex]. and [itex]\cos\pm\pi = -1[/itex].
I have some series left from evaluating at pi and I have exactly the same series evaluated at -pi. Is the integral equal to 0? EDIT: NO, I forgot the x terms.

Is there any, more elegant solution to this problem, though?

x^2014 is an even function. sin(x) is an odd function. What kind of function is the product?
 
  • #4
I'm not sure, going with my gut I would say it's an odd function, which means that F(-x) + F(x) = 0
The evaluation would then look like F(pi) - F(-pi) = 2F(pi)?
 
  • #5
nuuskur said:
I'm not sure, going with my gut I would say it's an odd function, which means that F(-x) + F(x) = 0
The evaluation would then look like F(pi) - F(-pi) = 2F(pi)?

Yes, it's an odd function. The integral of any odd function on an interval like [-a,a] that's symmetric around the origin is zero. You haven't seen this before?
 
  • #6
nuuskur said:
I'm not sure, going with my gut I would say it's an odd function, which means that F(-x) + F(x) = 0
The evaluation would then look like F(pi) - F(-pi) = 2F(pi)?

Not quite. If ##f(x)## is odd on ##[-L, L]## or ##(-L, L)##, then ##\int_{-L}^{L} f(x) \space dx = 0##.
 

What is the concept of integral evaluation?

Integral evaluation involves finding the value of a definite integral, which represents the area under a curve between two specified points on the x-axis.

What is the process for solving definite integrals?

The process for solving definite integrals involves using techniques such as substitution, integration by parts, and trigonometric identities to simplify the integrand and then evaluating the integral using the fundamental theorem of calculus.

What is the importance of the limits of integration in integral evaluation?

The limits of integration determine the boundaries of the area under the curve that is being evaluated. They are crucial in determining the correct value of the integral and must be carefully chosen based on the problem at hand.

How do you handle improper integrals in integral evaluation?

Improper integrals occur when one or both of the limits of integration are infinite or when the integrand is undefined at a point within the interval. In these cases, the integral must be broken into smaller, finite integrals and then evaluated using the proper techniques.

What is the significance of the number of iterations in integral evaluation?

The number of iterations refers to the number of subintervals used to approximate the area under the curve. Generally, the more iterations used, the more accurate the approximation will be. However, increasing the number of iterations also increases the computation time, so a balance must be struck between accuracy and efficiency.

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