Integration Questions

Question Number 27410 by math1967 last updated on 06/Jan/18

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{sinxdx}}{{x}} \\$$

Commented by math1967 last updated on 07/Jan/18

$${Thank}\:{you}\:{sir} \\$$

Commented by abdo imad last updated on 06/Jan/18

$${let}\:{introduce}\:{the}\:{fonction}\:{f}\left({t}\right)=\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}\:{e}^{−{tx}} {dx}\:{with}\:{t}\geqslant\mathrm{0} \\$$$${after}\:{verifying}\:{that}\:{f}\:{is}\:{derivable}\:{we}\:{have} \\$$$${f}^{'} \left({t}\right)=\:−\int_{\mathrm{0}} ^{\infty} {sinx}\:{e}^{−{tx}} {dx}=−{Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{\left({i}−{t}\right){x}} {dx}\right) \\$$$$=−\left[\:\frac{\mathrm{1}}{{i}−{t}}\:{e}^{\left({i}−{t}\right){x}} \:\right]_{{x}=\mathrm{0}} ^{{x}−>\propto} =\:\frac{\mathrm{1}}{{i}−{t}}\:=\frac{{i}+{t}}{−\mathrm{1}−{t}^{\mathrm{2}} }=−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:−\frac{{i}}{\mathrm{1}+{t}^{\mathrm{2}} } \\$$$$\Rightarrow{f}^{'} \left({t}\right)=\:−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\Rightarrow{f}\left({t}\right)=\lambda−{arctan}\left({t}\right){and}\:{due}\:{to}\:{f}\:{continue} \\$$$$\exists\:{M}>\mathrm{0}//{f}\left({t}\right)\leqslant{M}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} {dx}=_{{t}−>\propto{so}} \frac{{M}}{{t}}−>\mathrm{0} \\$$$${so}\:\lambda=\frac{\pi}{\mathrm{2}}\:{and}\:{f}\left({t}\right)=\frac{\pi}{\mathrm{2}}\:−{arctan}\left({t}\right) \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}={f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\:\:. \\$$$$\\$$