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Question Number 134795 by 0731619177 last updated on 07/Mar/21

Answered by EDWIN88 last updated on 07/Mar/21

 noting that the integrand is even function  I=∫_0 ^∞ ((sin (πx))/(x(1−x^2 ))) dx = (1/2)∫_(−∞) ^∞ ((sin (πx))/(x(1−x^2 ))) dx  I= (1/4)∫_(−∞) ^∞ ((2/x)−(1/(x+1))−(1/(x−1))).sin (πx) dx   I= (1/4).(2∫_(−∞) ^∞ ((sin (πx))/x)dx−∫_(−∞) ^∞ ((sin (πx))/(x+1))dx−∫_(−∞) ^∞ ((sin (πx))/(x−1))dx )  I=(1/4).(2∫_(−∞) ^∞ ((sin (πx))/x)dx−∫_(−∞) ^∞ ((sin (π(x−1)))/x)dx−∫_(−∞) ^∞ ((sin (π(x+1)))/x)dx)  I=(1/4)(2∫_(−∞) ^∞ ((sin (πx))/x)dx+∫_(−∞) ^∞ ((sin (πx))/x)dx+∫_(−∞) ^∞ ((sin (πx))/x)dx)  I=∫_(−∞) ^∞ ((sin (πx))/x)dx . let u = πx →x=(u/π)  I=∫_(−∞) ^∞ ((sin u)/(((u/π)))) ((du/π))=∫_(−∞) ^∞ ((sin u)/u) du = π

$$\:\mathrm{noting}\:\mathrm{that}\:\mathrm{the}\:\mathrm{integrand}\:\mathrm{is}\:\mathrm{even}\:\mathrm{function} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}\:\mathrm{dx} \\ $$$$\mathrm{I}=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{−\infty} ^{\infty} \left(\frac{\mathrm{2}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\right).\mathrm{sin}\:\left(\pi\mathrm{x}\right)\:\mathrm{dx}\: \\ $$$$\mathrm{I}=\:\frac{\mathrm{1}}{\mathrm{4}}.\left(\mathrm{2}\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}−\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}+\mathrm{1}}\mathrm{dx}−\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}−\mathrm{1}}\mathrm{dx}\:\right) \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{4}}.\left(\mathrm{2}\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}−\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\left(\mathrm{x}−\mathrm{1}\right)\right)}{\mathrm{x}}\mathrm{dx}−\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\left(\mathrm{x}+\mathrm{1}\right)\right)}{\mathrm{x}}\mathrm{dx}\right) \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}+\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}+\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\right) \\ $$$$\mathrm{I}=\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left(\pi\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:.\:\mathrm{let}\:\mathrm{u}\:=\:\pi\mathrm{x}\:\rightarrow\mathrm{x}=\frac{\mathrm{u}}{\pi} \\ $$$$\mathrm{I}=\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\mathrm{u}}{\left(\frac{\mathrm{u}}{\pi}\right)}\:\left(\frac{\mathrm{du}}{\pi}\right)=\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\mathrm{u}}{\mathrm{u}}\:\mathrm{du}\:=\:\pi \\ $$

Commented by liberty last updated on 07/Mar/21

  ausgezeichneter Herr Edwin

$$ \\ $$ausgezeichneter Herr Edwin

Commented by 0731619177 last updated on 07/Mar/21

  tanks

$$ \\ $$$${tanks} \\ $$

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