Question Number 26559 by abdo imad last updated on 26/Dec/17
$${let}\:{put}\:{F}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} \:\frac{{sint}}{{t}}\:{dt}\:\:\:{with}\:\:{x}\geqslant\mathrm{0} \\ $$$${we}\:{accept}\:{that}\:{F}\:{is}\:{class}\:{C}^{\mathrm{1}} \:{on}\:\left[\mathrm{0},\propto\left[\right.\right. \\ $$$${calculate}\:\:\frac{\partial{F}}{\partial{x}}\:\:{and}\:{find}\:{F}\left({x}\right) \\ $$$${then}\:\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sint}}{{t}}\:{dt} \\ $$
Commented by prakash jain last updated on 27/Dec/17
![(dF/dx)=∫_0 ^∞ (∂/∂x)(e^(−tx) ((sin t)/t))dt =−∫_0 ^∞ e^(−tx) sin tdt =[((e^(−tx) (xsin t+cos t))/(x^2 +1))]_(t=0) ^(t=∞) =−(1/(x^2 +1)) F(x)=−tan^(−1) x+C F(∞)=0 ⇒C=(π/2) F(x)=−tan^(−1) x+(π/2) F(0)=∫_0 ^∞ ((sin t)/t)dt=(π/2)](Q26636.png)
$$\frac{{dF}}{{dx}}=\int_{\mathrm{0}} ^{\infty} \frac{\partial}{\partial{x}}\left({e}^{−{tx}} \frac{\mathrm{sin}\:{t}}{{t}}\right){dt} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} \mathrm{sin}\:{tdt} \\ $$$$=\left[\frac{{e}^{−{tx}} \left({x}\mathrm{sin}\:{t}+\mathrm{cos}\:{t}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\right]_{{t}=\mathrm{0}} ^{{t}=\infty} \\ $$$$=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${F}\left({x}\right)=−\mathrm{tan}^{−\mathrm{1}} {x}+{C} \\ $$$${F}\left(\infty\right)=\mathrm{0} \\ $$$$\Rightarrow{C}=\frac{\pi}{\mathrm{2}} \\ $$$${F}\left({x}\right)=−\mathrm{tan}^{−\mathrm{1}} {x}+\frac{\pi}{\mathrm{2}} \\ $$$${F}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{t}}{{t}}{dt}=\frac{\pi}{\mathrm{2}} \\ $$