0-sin-x-x-dx-

Question Number 171117 by meetbhavsar25 last updated on 08/Jun/22

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{sin}\:{x}}{{x}}{dx}\:=\:\left(?\right) \\ $$

Answered by aleks041103 last updated on 08/Jun/22

f(z)=∫_0 ^∞ e^(−zx) ((sinx)/x)dx lim_(z→∞) f(z) = 0 f(0)=∫_0 ^∞ ((sinx)/x)dx but lim_(z→∞) f(z) − f(0) = ∫_0 ^∞ f ′(z)dz=−f(0) ⇒∫_0 ^∞ ((sinx)/x)dx=−∫_0 ^∞ f ′(z)dz f ′(z)=(d/dz)(e^(−zx) )((sinx)/x)=−xe^(−zx) ((sinx)/x)= =−e^(−zx) sinx ⇒∫_0 ^∞ ((sinx)/x)dx=∫_0 ^∞ (∫_0 ^∞ e^(−zx) sinxdx)dz ∫_0 ^∞ e^(−zx) sinxdx determinant ((,D,I),(+,(sinx),e^(−zx) ),(−,(cosx),(−(e^(−zx) /z))),(+,(−sinx),(e^(−zx) /z^2 ))) ∫_0 ^∞ e^(−zx) sinxdx=[−((e^(−zx) sinx)/z)−((e^(−zx) cosx)/z^2 )]_0 ^∞ −(1/z^2 )∫_0 ^∞ e^(−zx) sinxdx ⇒(1+(1/z^2 ))∫_0 ^∞ e^(−zx) sinxdx=(1/z^2 ) ⇒∫_0 ^∞ e^(−zx) sinxdx=(1/(1+z^2 ))=f ′ (z) ⇒∫_0 ^∞ ((sinx)/x)dx=∫_0 ^∞ (dz/(1+z^2 ))=arctg(z)∣_0 ^∞ =(π/2) ⇒∫_0 ^∞ ((sinx)/x)dx=(π/2)

$${f}\left({z}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{zx}} \frac{{sinx}}{{x}}{dx} \\ $$$$\underset{{z}\rightarrow\infty} {{lim}}\:{f}\left({z}\right)\:=\:\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx} \\ $$$${but} \\ $$$$\underset{{z}\rightarrow\infty} {{lim}}\:{f}\left({z}\right)\:−\:{f}\left(\mathrm{0}\right)\:=\:\int_{\mathrm{0}} ^{\infty} {f}\:'\left({z}\right){dz}=−{f}\left(\mathrm{0}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=−\int_{\mathrm{0}} ^{\infty} {f}\:'\left({z}\right){dz} \\ $$$${f}\:'\left({z}\right)=\frac{{d}}{{dz}}\left({e}^{−{zx}} \right)\frac{{sinx}}{{x}}=−{xe}^{−{zx}} \frac{{sinx}}{{x}}= \\ $$$$=−{e}^{−{zx}} {sinx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} {e}^{−{zx}} {sinxdx}\right){dz} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{zx}} {sinxdx} \\ $$$$\begin{array}{|c|c|c|c|}{}&\hline{{D}}&\hline{{I}}\\{+}&\hline{{sinx}}&\hline{{e}^{−{zx}} }\\{−}&\hline{{cosx}}&\hline{−\frac{{e}^{−{zx}} }{{z}}}\\{+}&\hline{−{sinx}}&\hline{\frac{{e}^{−{zx}} }{{z}^{\mathrm{2}} }}\\\hline\end{array} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{zx}} {sinxdx}=\left[−\frac{{e}^{−{zx}} {sinx}}{{z}}−\frac{{e}^{−{zx}} {cosx}}{{z}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{zx}} {sinxdx} \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)\int_{\mathrm{0}} ^{\infty} {e}^{−{zx}} {sinxdx}=\frac{\mathrm{1}}{{z}^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} {e}^{−{zx}} {sinxdx}=\frac{\mathrm{1}}{\mathrm{1}+{z}^{\mathrm{2}} }={f}\:'\:\left({z}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{dz}}{\mathrm{1}+{z}^{\mathrm{2}} }={arctg}\left({z}\right)\mid_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}{dx}=\frac{\pi}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 09/Jun/22