A sinusoid integrated from -infinity to infinity

[pennyantics]

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.

 

[Gib Z]

Now my initial guess because of the antisymmetry of the function is that it equals zero.

Exactly correct.

 

[christianjb]

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.

 

[Gib Z]

$$\lim_{a\to\infty}( \int^a_{-a} \sin^3 x dx ) = \lim_{a\to\infty} (\int^a_0 \sin^3 x dx + \int^0_{-a} \sin^3 x dx) =\lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(-a))$$, Where dF(x)/dx=sin^3 x.

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

$$\lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(-a)) = \lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(a)) =\lim_{a\to\infty} (0) = 0$$.

 

[HallsofIvy]

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