- #1

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[itex]... + \int_0^{\infty}x^rf_1(x)sin(2\pi logx)dx[/itex].

*However, the transformation [itex]y=-logx-r[/itex] shows that this last integral is that of an odd function over (-∞,∞) and hence equal to 0 for [itex]r=0,1,...[/itex]*

By the way, the author means the natural log. I find this annoying.

Now assuming that the integrand is an odd function, I'm not so sure that this last statement is true. Is it the case that the integral of an odd function over (-∞,∞) equals 0.

Unless I've made a mistake (which happens often), I have a counterexample. The function, sin(x) is an odd function. Then

[itex]\int_{-\infty}^{\infty}sin(x)dx = \int_{-\infty}^0sin(x)dx + \int_0^{\infty}sin(x)dx[/itex].

Then looking at the second integral of the right-hand-side, we have

[itex]lim_{t \to \infty}\int_0^{t}sin(x)dx = lim_{t \to \infty} -cos(x) |_0^{t} [/itex],

but [itex]lim_{t \to \infty}-cos(x) [/itex] does not exist.

So even though sin(x) is an odd function, the integral of sin(x) over (-∞,∞) diverges, so it does not equal 0.

Am I right here?