Q1146 (Q1146): Difference between revisions
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Created claim: formatter url (P4): \int_0^\infty\frac{|\sin(x)|}x~\mathrm dx&=\sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}x~\mathrm dx\\&=\sum_{n=0}^\infty\int_0^\pi\frac{\sin(x)}{x+n\pi}~\mathrm dx\\&=\sum_{n=0}^\infty\left[\frac1{(n+1)\pi}+\frac1{n\pi}-\int_0^\pi\frac{\cos(x)}{(x+n\pi)^2}~\mathrm dx\right]\tag{ibp} |
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| Property / math stackexcange post id | |||
| Property / math stackexcange post id: 3229015 / rank | |||
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| Property / relevant answers | |||
| Property / relevant answers: 3229032 / rank | |||
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| Property / relevant answers: 3229032 / qualifier | |||
Defining formula: see formatterurl | |||
| Property / formatter url | |||
\int_0^\infty\frac{|\sin(x)|}x~\mathrm dx&=\sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}x~\mathrm dx\\&=\sum_{n=0}^\infty\int_0^\pi\frac{\sin(x)}{x+n\pi}~\mathrm dx\\&=\sum_{n=0}^\infty\left[\frac1{(n+1)\pi}+\frac1{n\pi}-\int_0^\pi\frac{\cos(x)}{(x+n\pi)^2}~\mathrm dx\right]\tag{ibp} | |||
| Property / formatter url: \int_0^\infty\frac{|\sin(x)|}x~\mathrm dx&=\sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}x~\mathrm dx\\&=\sum_{n=0}^\infty\int_0^\pi\frac{\sin(x)}{x+n\pi}~\mathrm dx\\&=\sum_{n=0}^\infty\left[\frac1{(n+1)\pi}+\frac1{n\pi}-\int_0^\pi\frac{\cos(x)}{(x+n\pi)^2}~\mathrm dx\right]\tag{ibp} / rank | |||
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Latest revision as of 20:59, 19 May 2020
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Statements
B.26
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q_205
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\int_0^\infty\frac{|\sin(x)|}x~\mathrm dx&=\sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}x~\mathrm dx\\&=\sum_{n=0}^\infty\int_0^\pi\frac{\sin(x)}{x+n\pi}~\mathrm dx\\&=\sum_{n=0}^\infty\left[\frac1{(n+1)\pi}+\frac1{n\pi}-\int_0^\pi\frac{\cos(x)}{(x+n\pi)^2}~\mathrm dx\right]\tag{ibp}
0 references