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I've come up with the following proof for $\int^{6\pi}_{4\pi} \frac{\sin x}{x}dx = 0$:

The statement is true, due to the mean value theorem specifying the existence of a $k \in [2\pi,4\pi]$ such that:

\begin{align*} \int^{6\pi}_{4\pi} \frac{\sin x}{x}dx &= \frac{1}{k}\int^{6\pi}_{4\pi}\sin x dx \\ &= \frac{1}{k}(- \cos x \bigg|^{6\pi}_{4\pi}) \\ &= \frac{\cos 4\pi - \cos 6\pi}{k} \\ &= 0 \end{align*}

However according to Wolfram Alpha this does not seem to be correct. Is the mean value theorem not applicable here for some reason, or did I do something else wrong here?

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    $\begingroup$ You can't apply mean value theorem for definite integral here because $\sin x$ is neither non-negative nor non-positive over $[4\pi,6\pi]$. $\endgroup$ – achille hui Nov 7 '16 at 2:57
  • $\begingroup$ I do not see how what you wrote relates to the mean value theorem. Could you please include the statement of the MVT, and how you apply it. $\endgroup$ – Mirko Nov 7 '16 at 3:06
  • $\begingroup$ @Mirko I'm using the MVT $\int_a^b f(x) g(x) dx = f(c) \int_a^b g(x) dx$ with $f(x) = \frac{1}{x}$ and $g(x) = \sin(x)$. $\endgroup$ – ntldr Nov 7 '16 at 3:11
  • $\begingroup$ aah, well, but then @achillehui already observed that in this case $g$ need to have the same sign on $[a,b]$ en.wikipedia.org/wiki/… $\endgroup$ – Mirko Nov 7 '16 at 3:14
  • $\begingroup$ Oh yes... of cause. Thanks a lot! @achillehui could you please post your comment as answer, so I can accept it? $\endgroup$ – ntldr Nov 7 '16 at 3:20
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One version of mean value theorem for definite integral states:

Given two functions $f, g$ on $[ a, b ]$. If $f$ is continuous and $g$ is an integrable function which doesn't change sign on $[ a, b ]$, then there exists $c \in (a,b)$ such that $$\int_a^b f(x) g(x) dx = f(c) \int_a^b g(x) dx$$

For your case, you can't apply this version of mean value theorem because $g(x) = \sin x$ changes sign over $[4\pi, 6\pi]$.

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$$\int\frac{\sin x}x\,dx$$ does not evaluate to $-\cos x+K$ because of the division by $x$. This is one of the trigonometric integrals and has its own notation: $$\int\frac{\sin x}x\,dx=\operatorname{Si}(x)+K$$ Therefore your proof is wrong and $$\int_{4\pi}^{6\pi}\frac{\sin x}x\,dx=\operatorname{Si}(6\pi)-\operatorname{Si}(4\pi)=0.02587\dots$$

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