Question Number 133563 by bemath last updated on 23/Feb/21
![∫_0 ^∞ [sin (1/x)−(1/π)sin ((π/x))] dx](https://www.tinkutara.com/question/Q133563.png)
$$\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\left[\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\pi}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{x}}\right)\right]\:\mathrm{dx} \\ $$
Answered by EDWIN88 last updated on 23/Feb/21
![Calculate ∫_0 ^∞ (sin( (1/x))−(1/π) sin ((π/x) ) )dx let (1/x) = t∧ x= (1/t) ; determinant (((x→∞),(t→0^+ )),((x→0^+ ),(t→∞))) I=∫_∞ ^0 (sin t−(1/π)sin πt)(−(1/t^2 ))dt I=∫_0 ^∞ (((((sin t)/t)− ((sin πt)/(πt)))/t) ) dt ; Frullani integral I=[ lim_(t→∞) (((sin t)/t))−lim_(t→0) ((sin πt)/(πt)) ].ln ((b/a)) ; { ((b=1)),((a=π)) :} I=(0−1)ln ((1/π))=ln π](https://www.tinkutara.com/question/Q133567.png)
$$\:\mathrm{Calculate}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\mathrm{sin}\left(\:\frac{\mathrm{1}}{\mathrm{x}}\right)−\frac{\mathrm{1}}{\pi}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{x}}\:\right)\:\right)\mathrm{dx} \\ $$$$\mathrm{let}\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{t}\wedge\:\mathrm{x}=\:\frac{\mathrm{1}}{\mathrm{t}}\:;\:\begin{array}{|c|c|}{\mathrm{x}\rightarrow\infty}&\hline{\mathrm{t}\rightarrow\mathrm{0}^{+} }\\{\mathrm{x}\rightarrow\mathrm{0}^{+} }&\hline{\mathrm{t}\rightarrow\infty}\\\hline\end{array} \\ $$$$\mathrm{I}=\underset{\infty} {\overset{\mathrm{0}} {\int}}\left(\mathrm{sin}\:\mathrm{t}−\frac{\mathrm{1}}{\pi}\mathrm{sin}\:\pi\mathrm{t}\right)\left(−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt}\: \\ $$$$\mathrm{I}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\frac{\mathrm{sin}\:\mathrm{t}}{\mathrm{t}}−\:\frac{\mathrm{sin}\:\pi\mathrm{t}}{\pi\mathrm{t}}}{\mathrm{t}}\:\right)\:\mathrm{dt}\:;\:\mathrm{Frullani}\:\mathrm{integral} \\ $$$$\mathrm{I}=\left[\:\underset{\mathrm{t}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\mathrm{t}}{\mathrm{t}}\right)−\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\pi\mathrm{t}}{\pi\mathrm{t}}\:\right].\mathrm{ln}\:\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\:;\:\begin{cases}{\mathrm{b}=\mathrm{1}}\\{\mathrm{a}=\pi}\end{cases} \\ $$$$\mathrm{I}=\left(\mathrm{0}−\mathrm{1}\right)\mathrm{ln}\:\left(\frac{\mathrm{1}}{\pi}\right)=\mathrm{ln}\:\pi\: \\ $$$$ \\ $$