Question Number 27410 by math1967 last updated on 06/Jan/18
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{sinxdx}}{{x}} \\ $$
Commented bymath1967 last updated on 07/Jan/18
$${Thank}\:{you}\:{sir} \\ $$
Commented byabdo imad last updated on 06/Jan/18
![let introduce the fonction f(t)= ∫_0 ^∞ ((sinx)/x) e^(−tx) dx with t≥0 after verifying that f is derivable we have f^′ (t)= −∫_0 ^∞ sinx e^(−tx) dx=−Im(∫_0 ^∞ e^((i−t)x) dx) =−[ (1/(i−t)) e^((i−t)x) ]_(x=0) ^(x−>∝) = (1/(i−t)) =((i+t)/(−1−t^2 ))=−(t/(1+t^2 )) −(i/(1+t^2 )) ⇒f^′ (t)= −(1/(1+t^2 )) ⇒f(t)=λ−arctan(t)and due to f continue ∃ M>0//f(t)≤M∫_0 ^∞ e^(−tx) dx=_(t−>∝so) (M/t)−>0 so λ=(π/2) and f(t)=(π/2) −arctan(t) ∫_0 ^∞ ((sinx)/x)=f(0)=(π/2) .](Q27433.png)
$${let}\:{introduce}\:{the}\:{fonction}\:{f}\left({t}\right)=\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}\:{e}^{−{tx}} {dx}\:{with}\:{t}\geqslant\mathrm{0} \\ $$ $${after}\:{verifying}\:{that}\:{f}\:{is}\:{derivable}\:{we}\:{have} \\ $$ $${f}^{'} \left({t}\right)=\:−\int_{\mathrm{0}} ^{\infty} {sinx}\:{e}^{−{tx}} {dx}=−{Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{\left({i}−{t}\right){x}} {dx}\right) \\ $$ $$=−\left[\:\frac{\mathrm{1}}{{i}−{t}}\:{e}^{\left({i}−{t}\right){x}} \:\right]_{{x}=\mathrm{0}} ^{{x}−>\propto} =\:\frac{\mathrm{1}}{{i}−{t}}\:=\frac{{i}+{t}}{−\mathrm{1}−{t}^{\mathrm{2}} }=−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:−\frac{{i}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$ $$\Rightarrow{f}^{'} \left({t}\right)=\:−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\Rightarrow{f}\left({t}\right)=\lambda−{arctan}\left({t}\right){and}\:{due}\:{to}\:{f}\:{continue} \\ $$ $$\exists\:{M}>\mathrm{0}//{f}\left({t}\right)\leqslant{M}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} {dx}=_{{t}−>\propto{so}} \frac{{M}}{{t}}−>\mathrm{0} \\ $$ $${so}\:\lambda=\frac{\pi}{\mathrm{2}}\:{and}\:{f}\left({t}\right)=\frac{\pi}{\mathrm{2}}\:−{arctan}\left({t}\right) \\ $$ $$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}={f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\:\:. \\ $$ $$ \\ $$