Integration over a discontinuity

[nrivera1]

Homework Statement

Evaluate ∫(1/x)sin^2(x)dx from -a to a

The Attempt at a Solution

Mathematica doesn't want to evaluate this because of the lack of convergence.

I think it is zero. When we consider non-zero values of x in the associated riemann sum, the integrand is odd and so contributions from x and -x cancel out. I can make the Δx in the riemann sum arbitrarily small, so when we consider the step near zero, sin^2 x is almost exactly equal to x^2 and that term in the sum gets replaced to xΔx where x is very nearly zero as is delta x. In other words, the integral is zero.

Is this kind of reasoning valid?

 

[antibrane]

Yea, it is zero by symmetry. The integrand is odd (it obeys $f(-x)=-f(x)$) so the "negative area" accumulated from -a to 0 cancels out the "positive area" accumulated from 0 to a.

 

[jbunniii]

Homework Statement

Evaluate ∫(1/x)sin^2(x)dx from -a to a

The Attempt at a Solution

Mathematica doesn't want to evaluate this because of the lack of convergence.

I think it is zero. When we consider non-zero values of x in the associated riemann sum, the integrand is odd and so contributions from x and -x cancel out. I can make the Δx in the riemann sum arbitrarily small, so when we consider the step near zero, sin^2 x is almost exactly equal to x^2 and that term in the sum gets replaced to xΔx where x is very nearly zero as is delta x. In other words, the integral is zero.

Is this kind of reasoning valid?

Yes, your reasoning is OK. More formally, we may write the integrand as
$$\frac{\sin^2(x)}{x} = \left(\frac{\sin(x)}{x}\right)\sin(x)$$
Both factors on the right hand side are continuous on [-a,a] (we continuously extend $\sin(x)/x$ to equal 1 at $x = 0$), so their product is also continuous, hence integrable. Therefore, since the product is an odd function and we are integrating over [-a,a], the result is zero.

 

[lurflurf]

since f(0^-)=f(0⁺)=0

this is no more of a problem than x^2/x would be

in fact x~sin(x) near zero

 

[haruspex]

Yea, it is zero by symmetry. The integrand is odd (it obeys $f(-x)=-f(x)$) so the "negative area" accumulated from -a to 0 cancels out the "positive area" accumulated from 0 to a.

That's not enough by itself. You also have to show that the integral is bounded on all subintervals, otherwise the symmetry argument might equate to saying ∞-∞ = 0. Other posts in this thread cover that.