Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com