# Is it ok to claim that an integral of an odd function in a symmetric interval is 0?

For instance,

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

Since sinx is odd, and the interval is symmetrical?

• Yes, this is perfectly valid. – greelious Dec 29 '18 at 18:48
• Yes, as long as it's convergent – Jakobian Dec 29 '18 at 18:50
• So you mean $$\int_{-1}^1\sin(x)(7-\cos^4(x))dx$$? – Dr. Sonnhard Graubner Dec 29 '18 at 18:53
• @Dr.SonnhardGraubner Funnily enough it isn't relevant to the question. – 0x539 Dec 29 '18 at 19:11

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*}
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.
• @Mustang AFAIK these functions are not Lebesgue integrable since the integral of their positive/negative parts is $\pm \infty$ – 0x539 Dec 29 '18 at 19:07
• 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? – Mustang Dec 29 '18 at 19:14