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For instance,

$$\int_{-1}^1 \sin x(7-\cos^4x) \,dx = 0$$

Since sinx is odd, and the interval is symmetrical?

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    $\begingroup$ Yes, this is perfectly valid. $\endgroup$ – greelious Dec 29 '18 at 18:48
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    $\begingroup$ Yes, as long as it's convergent $\endgroup$ – Jakobian Dec 29 '18 at 18:50
  • $\begingroup$ So you mean $$\int_{-1}^1\sin(x)(7-\cos^4(x))dx$$? $\endgroup$ – Dr. Sonnhard Graubner Dec 29 '18 at 18:53
  • $\begingroup$ @Dr.SonnhardGraubner Funnily enough it isn't relevant to the question. $\endgroup$ – 0x539 Dec 29 '18 at 19:11
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Yes this is perfectly legal. In the case, if $f(x)$ is odd, then from $-a$ to $a$, the integral vanishes. This can be proven by splitting up the integral

$$\begin{align*}\int\limits_{-a}^{a}\mathrm dx\, f(x) &=\int\limits_{0}^{a}\mathrm dx\, f(x)+\int\limits_{-a}^0\mathrm dx\,f(x)\\ & =\int\limits_{0}^a\mathrm dx\, f(x)-\int\limits_{0}^a\mathrm dx\, f(x)\\ & =0\end{align*}$$

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Basically yes. Note however that integrals like $$\int_{-1}^1 \frac1{x} \mathrm{d} x \quad\text{or}\quad \int_{-\infty}^\infty \frac{x}{1 + x^2} \mathrm{d}x$$ cannot be said to be zero because they fail to converge.

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  • $\begingroup$ Yes, it is true for riemann definition of integral, but in the lebegue definition, you can in a way ignore the singular point and calculate the integral. Here again, it would be zero. There is some discussion on this here 'math.stackexchange.com/a/3048916/584828' if you're interested. $\endgroup$ – Mustang Dec 29 '18 at 19:00
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    $\begingroup$ @Mustang AFAIK these functions are not Lebesgue integrable since the integral of their positive/negative parts is $\pm \infty$ $\endgroup$ – 0x539 Dec 29 '18 at 19:07
  • $\begingroup$ Sorry, I'm not too clear about the details myself(I am not familiar with lebegue theory), this is basically what my prof. told me when I asked him the same question(for $\int_{-1}^{1}\frac{1}{x}$ he said we could take $x=0$ as the principal value and work it out). This is also roughly what one guy answered in the link. Hopefully someone more experienced can shed some light here? $\endgroup$ – Mustang Dec 29 '18 at 19:14
  • $\begingroup$ I say - don't confuse things. If the integral exists, it's zero. A principal value is a way to stretch that "exists" to something broader, but it's not part of the standard definition. As for Lebesgue integrals - no. Principal values are not a thing in the Lebesgue theory. $\endgroup$ – jmerry Dec 29 '18 at 21:15

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