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Question Number 46502 by Ac Kesharwani last updated on 27/Oct/18

$$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{sin}\:\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\right)\boldsymbol{\mathrm{dx}} \\$$

Commented by math khazana by abdo last updated on 27/Oct/18

$${let}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{{x}}{e}^{−{tx}} {dx}\:{with}\:{t}\geqslant\mathrm{0}\:{we}\:{can}\:{prove}\: \\$$$${that}\:{f}\:{is}\:{continue}\:{derivable}\:{on}\:\left[\mathrm{0},+\infty\left[\:{and}\:{we}\:{have}\right.\right. \\$$$${f}^{'} \left({t}\right)=−\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} {sinx}\:{dx}\:=−{Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}+{ix}} {dx}\right)\:{but} \\$$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{t}+{i}\right){x}} {dx}=\left[\frac{\mathrm{1}}{−{t}+{i}}\:{e}^{\left(−{t}+{i}\right){x}} \right]_{{x}=\mathrm{0}} ^{+\infty} =\frac{−\mathrm{1}}{−{t}+{i}} \\$$$$=\frac{\mathrm{1}}{{t}−{i}}=\frac{{t}+{i}}{{t}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow{f}^{'} \left({t}\right)=−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow \\$$$${f}\left({t}\right)=−{arctan}\left({t}\right)+\lambda\:\:{but}\:\exists{m}>\mathrm{0}\:/ \\$$$$\mid{f}\left({t}\right)\mid\leqslant{m}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} {dx}=\frac{{m}}{{t}}\rightarrow\mathrm{0}\:\left({t}\rightarrow+\infty\right)\Rightarrow\lambda=\frac{\pi}{\mathrm{2}} \\$$$${f}\left({t}\right)=\frac{\pi}{\mathrm{2}}\:−{arctan}\left({t}\right)\:\Rightarrow{f}\left({o}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=\frac{\pi}{\mathrm{2}} \\$$$$\left({arctan}\mathrm{0}=\mathrm{0}\right) \\$$

Answered by MJS last updated on 27/Oct/18

$$\int\frac{\mathrm{sin}\:{x}}{{x}}{dx}=\mathrm{Si}\left({x}\right)\:\left[\mathrm{integral}−\mathrm{sinus},\:\mathrm{table}\:\mathrm{of}\right. \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{values}\:\mathrm{should}\:\mathrm{exist}\:\mathrm{somewhere} \\$$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{in}\:\mathrm{the}\:\mathrm{www}\right] \\$$$$\mathrm{Si}\left(\infty\right)−\mathrm{Si}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}} \\$$