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Why is $$\int_{0}^{\infty}{\,\frac{1\,-\,\cos{y}}{y^2}\,dy}=\int_{0}^{\infty}{\,\frac{\sin{y}}{y}\,dy}?$$

It should be solved with integration by parts.

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  • $\begingroup$ Let $u=1-\cos y$ and $dv=(1/y^2)dy$. Work it out ... $\endgroup$ – David Mitra Jan 4 '15 at 13:15
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So, I got with integration by parts and with $f(y)=-\frac{1}{y}$, $f'(y)=\frac{1}{y^2}$, $g(y)=1-\cos(y)$ and $g'(y)=\sin(y)$ $$ \int_{0}^{\infty}{\frac{1}{y^2}\cdot(1-\cos{y})dy}\,=\,\Big[-\frac{1}{y}\cdot(1-\cos(y))\Big]_0^{\infty}-\int_{0}^{\infty}{\Big[-\frac{1}{y}\cdot(\sin{y})\Big]dy}\,=\quad-\int_{0}^{\infty}{(-1)\cdot\frac{\sin(y)}{y}dy}\,=\,\int_{0}^{\infty}{\frac{\sin(y)}{y}dy}. $$ Because $\lim\limits_{x\to0^+}{\frac{1-\cos x}{x}}=0$ and $\lim\limits_{x\to\infty}{1-\cos x\over x}=0$.

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  • $\begingroup$ Looks good. You should provide details showing that the non-integral term after applying the integration by parts formula is indeed $0$, though. $\endgroup$ – David Mitra Jan 4 '15 at 13:51
  • $\begingroup$ No, that doesn't quite justify it. You need to evaluate $\lim_{x\rightarrow0^+}{1-\cos x\over x}$ (at the endpoint $\infty$, the limit is clearly $0$). $\endgroup$ – David Mitra Jan 4 '15 at 13:53
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Primitive function of $1/y^2$ is $-1/y$ and derivative of $1-\cos x$ is $\sin x$. You do the work with limits...

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Hint: $\dfrac{1-\cos(2t)}2=\sin^2t$.

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