What's a simple way to display $$\int_1^{\infty} \frac{\sin(x)}{x}dx$$ conditionally convergent (i.e. convergent, but not absolutely)?
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1Integration by parts, with $u=\frac{1}{x}$. – Dustan Levenstein May 15 '16 at 13:34
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To show it isn't absolutely convergent, maybe you could approximate it by triangular areas beneath and use a comparison test. That may be algebraically complicated, but conceptually simple. – jdods May 15 '16 at 13:34
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For that part I think I would just use the fact that $|sin(x)| \ge \frac{1}{2}$ periodically often. – Dustan Levenstein May 15 '16 at 13:40
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@Did It's clearly not the same question? – mavavilj May 15 '16 at 14:42
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This one clearly has an answer over there: don't you read the answers you receive? – Did May 15 '16 at 14:43
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The interval there is $\int_0^{\infty}$. – mavavilj May 15 '16 at 14:44
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1What? $ $ $ $ $ $ – Did May 15 '16 at 15:33
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Dirichlet's test: $\sin x$ has a bounded primitive and $\frac{1}{x}$ is decreasing to zero, hence $\int_{1}^{+\infty}\frac{\sin x}{x}\,dx $ is conditionally convergent. It cannot be absolutely convergent since the mean value of $|\sin x|$ is $\frac{2}{\pi}$, so $\int_{1}^{M}\frac{|\sin x|}{x}\,dx $ behaves like $\frac{2}{\pi}\log M$ by integration by parts.
Jack D'Aurizio
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