Question Number 46091 by sandeepkeshari0797@gmail.com last updated on 21/Oct/18
Commented by maxmathsup by imad last updated on 21/Oct/18
![let f(t)=∫_0 ^u ((sinx)/x)e^(−tx) dx with t≥o we have f^′ (t)=−∫_0 ^u sinx e^(−tx) dx ⇒−f^′ (t)=Im(∫_0 ^u e^(ix) e^(−tx) dx)=Im ( ∫_0 ^u e^((i−t)x) dx) but ∫_0 ^u e^((i−t)x) dx =[(1/(i−t)) e^((i−t)x) ]_0 ^u =(1/(i−t))( e^((i−t)u) −1) = (1/(t−i))( 1−e^(−tu) (cosu +isinu))=((t+i)/(t^2 +1)){ 1−e^(−tu) cosu −i e^(−tu) sinu} =((t−t e^(−tu) cosu−it e^(−tu) sinu +i(1−e^(−tu) cosu )+e^(−tu) sinu)/(1+t^2 )) =((t−t e^(−tu) cosu +e^(−tu) sinu +i(1−e^(−tu) cosu−t e^(−tu) sinu))/(1+t^2 )) ⇒ f^′ (t)=((t e^(−tu) sinu +e^(−tu) cosu −1)/(1+t^2 )) ⇒ f(t)= ∫_0 ^t ((x e^(−xu) sinu +e^(−xu) cosu−1)/(1+x^2 ))dx+c c=f(o)=∫_0 ^u ((sinx)/x)dx ....be continued...](https://www.tinkutara.com/question/Q46112.png)
$${let}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{{u}} \:\frac{{sinx}}{{x}}{e}^{−{tx}} {dx}\:\:{with}\:{t}\geqslant{o}\:\:{we}\:{have}\:{f}^{'} \left({t}\right)=−\int_{\mathrm{0}} ^{{u}} \:{sinx}\:{e}^{−{tx}} {dx} \\ $$$$\Rightarrow−{f}^{'} \left({t}\right)={Im}\left(\int_{\mathrm{0}} ^{{u}} \:{e}^{{ix}} \:{e}^{−{tx}} {dx}\right)={Im}\:\left(\:\int_{\mathrm{0}} ^{{u}} \:\:{e}^{\left({i}−{t}\right){x}} {dx}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{{u}} \:{e}^{\left({i}−{t}\right){x}} {dx}\:=\left[\frac{\mathrm{1}}{{i}−{t}}\:{e}^{\left({i}−{t}\right){x}} \right]_{\mathrm{0}} ^{{u}} \:=\frac{\mathrm{1}}{{i}−{t}}\left(\:{e}^{\left({i}−{t}\right){u}} −\mathrm{1}\right) \\ $$$$=\:\frac{\mathrm{1}}{{t}−{i}}\left(\:\mathrm{1}−{e}^{−{tu}} \left({cosu}\:+{isinu}\right)\right)=\frac{{t}+{i}}{{t}^{\mathrm{2}} +\mathrm{1}}\left\{\:\mathrm{1}−{e}^{−{tu}} {cosu}\:−{i}\:{e}^{−{tu}} \:{sinu}\right\} \\ $$$$=\frac{{t}−{t}\:{e}^{−{tu}} {cosu}−{it}\:{e}^{−{tu}} {sinu}\:+{i}\left(\mathrm{1}−{e}^{−{tu}} {cosu}\:\right)+{e}^{−{tu}} {sinu}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\frac{{t}−{t}\:{e}^{−{tu}} {cosu}\:+{e}^{−{tu}} {sinu}\:\:+{i}\left(\mathrm{1}−{e}^{−{tu}} {cosu}−{t}\:{e}^{−{tu}} {sinu}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)=\frac{{t}\:{e}^{−{tu}} {sinu}\:+{e}^{−{tu}} \:{cosu}\:−\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow\:{f}\left({t}\right)=\:\int_{\mathrm{0}} ^{{t}} \:\:\frac{{x}\:{e}^{−{xu}} {sinu}\:+{e}^{−{xu}} {cosu}−\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}+{c} \\ $$$${c}={f}\left({o}\right)=\int_{\mathrm{0}} ^{{u}} \:\frac{{sinx}}{{x}}{dx}\:\:\:….{be}\:{continued}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18
$${sinx}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+… \\ $$$$\int\frac{{sinx}}{{x}}{dx} \\ $$$$\int\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{7}!}+…\:{dx} \\ $$$$={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}×\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}×\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}×\mathrm{7}!}+…\:\:\:\:\:\:{C} \\ $$