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A sinusoid integrated from -infinity to infinity

  • #1
I had a sort of odd question on my homework,

Sin(x)^3 dx, integrated over all reals (from negative infinity to infinity).

The problem also gives this morsel of ambiguity:

"Hint: think before integrating. this is easy"

Now my initial guess because of the antisymmetry of the function is that it equals zero. Although the problem doesn't ask for a proof of any way shape or form however, I was baffled how I would argue that I reasoned it equaled zero if I was called upon in class.

So I'm wondering whether my assumption is correct as well as maybe a brief explanation. No proof needed.
 

Answers and Replies

  • #2
Gib Z
Homework Helper
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5
Now my initial guess because of the antisymmetry of the function is that it equals zero.
Exactly correct.
 
  • #3
529
1
A pedant might ask for proof that you can use the antisymmetry of the integral in this way when the limits are +- infinity.

But I guess that's why we have mathematicians.
 
  • #4
Gib Z
Homework Helper
3,346
5
[tex]\lim_{a\to\infty}( \int^a_{-a} \sin^3 x dx )

= \lim_{a\to\infty} (\int^a_0 \sin^3 x dx + \int^0_{-a} \sin^3 x dx)

=\lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(-a))[/tex], Where dF(x)/dx=sin^3 x.

Since the derivative of any odd function is an even function, F(-a)=F(a)

[tex]\lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(-a)) = \lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(a))

=\lim_{a\to\infty} (0)

= 0[/tex].
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,806
932
What Gib Z gives is the "Cauchy Principal Value" of the integral. Of course, the limit is 0 because sin(x) is an odd function. Evaluating its integral at a and -a will give the same thing.

Strictly speaking [itex]\int_{-\infty}^\infty f(x)dx[/itex] is
[tex]\lim_{a\rightarrow -\infty}\int_a^0 f(x)dx+ \lim_{b\rightarrow \infty} \int_0^b f(x) dx[/tex]
where the two limits are taken independently. Using that definition,
[tex]\int_{-\infty}^\infty sin^3(x) dx[/tex]
does not exist.
 

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