Menu Close

Question-102985

Tinku Tara 0 Comments
Pin




Question Number 102985 by DGmichael last updated on 12/Jul/20
Answered by mathmax by abdo last updated on 12/Jul/20
at form of serie let f(x) =∫_0 ^x (e^t /((1+t^2 )^2 ))dt if o≤x<1 we have (1/(1+u)) =Σ_(n=0) ^∞ (−1)^n u^(n ) ⇒by derivation we get −(1/((1+u)^2 )) =Σ_(n=1) ^∞ n(−1)^n u^(n−1) ⇒(1/((1+u)^2 )) =Σ_(n=1) ^∞ n(−1)^(n−1) u^(n−1) ⇒ (1/((1+t^2 )^2 )) =Σ_(n=1) ^∞ n(−1)^(n−1) t^(2n−2) ⇒f(x) =∫_0 ^(x ) e^t (Σ_(n=1) ^(∞ ) n(−1)^(n−1 ) t^(2n−2) )dt =Σ_(n=1) ^∞ n(−1)^(n−1) ∫_0 ^(x ) t^(2n−2) e^t dt =Σ_(n=1) ^∞ n(−1)^(n−1) U_n U_n =∫_0 ^x t^(2n−2 ) e^t dt by parts u^′ =t^(2n−2) and v =e^t ⇒ U_n =[(1/(2n−1)) t^(2n−1) e^t ]_0 ^x −∫_0 ^x (1/(2n−1)) t^(2n−1) e^(t ) dt =((x^(2n−1) e^x )/(2n−1)) −(1/(2n−1)) ∫_0 ^x t^(2n−1) e^t dt and ∫_0 ^x t^(2n−1) e^t dt =[(t^(2n) /(2n)) e^t ]_0 ^x −∫_0 ^x (t^(2n) /(2n))e^t dt =((x^(2n) e^x )/(2n)) −(1/(2n)) ∫_0 ^x t^(2n) e^t dt =((x^(2n) e^x )/(2n))−(1/(2n))U_(n+1) ⇒U_n =((x^(2n−1) e^x )/(2n−1))−(1/(2n−1)){((x^(2n) e^x )/(2n))−(1/(2n)) U_(n+1) } =((x^(2n−1) e^x )/(2n−1))−((x^(2n) e^x )/((2n−1)(2n))) +(1/(2n(2n−1))) U_(n+1) ...be continued
$$\mathrm{at}\:\mathrm{form}\:\mathrm{of}\:\mathrm{serie}\:\:\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{e}^{\mathrm{t}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathrm{if}\:\mathrm{o}\leqslant\mathrm{x}<\mathrm{1}\:\:\mathrm{we}\:\:\mathrm{have}\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}\:} \:\Rightarrow\mathrm{by}\:\mathrm{derivation}\:\mathrm{we}\:\mathrm{get} \\ $$$$−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{n}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}−\mathrm{1}} \:\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{n}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{u}^{\mathrm{n}−\mathrm{1}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{n}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{t}^{\mathrm{2n}−\mathrm{2}} \:\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{x}\:} \mathrm{e}^{\mathrm{t}} \left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty\:} \mathrm{n}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}\:} \mathrm{t}^{\mathrm{2n}−\mathrm{2}} \right)\mathrm{dt} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{n}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\mathrm{x}\:} \mathrm{t}^{\mathrm{2n}−\mathrm{2}} \:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{n}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{U}_{\mathrm{n}} \\ $$$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{x}} \:\mathrm{t}^{\mathrm{2n}−\mathrm{2}\:} \:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:\:\:\mathrm{by}\:\mathrm{parts}\:\:\mathrm{u}^{'} \:=\mathrm{t}^{\mathrm{2n}−\mathrm{2}} \:\mathrm{and}\:\mathrm{v}\:=\mathrm{e}^{\mathrm{t}} \:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} =\left[\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}\:\mathrm{t}^{\mathrm{2n}−\mathrm{1}} \:\mathrm{e}^{\mathrm{t}} \right]_{\mathrm{0}} ^{\mathrm{x}} \:−\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}\:\mathrm{t}^{\mathrm{2n}−\mathrm{1}} \:\mathrm{e}^{\mathrm{t}\:} \mathrm{dt} \\ $$$$=\frac{\mathrm{x}^{\mathrm{2n}−\mathrm{1}} \:\mathrm{e}^{\mathrm{x}} }{\mathrm{2n}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{x}} \:\mathrm{t}^{\mathrm{2n}−\mathrm{1}} \:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:\mathrm{and} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{x}} \:\mathrm{t}^{\mathrm{2n}−\mathrm{1}} \:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:=\left[\frac{\mathrm{t}^{\mathrm{2n}} }{\mathrm{2n}}\:\mathrm{e}^{\mathrm{t}} \right]_{\mathrm{0}} ^{\mathrm{x}} \:−\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{t}^{\mathrm{2n}} }{\mathrm{2n}}\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:=\frac{\mathrm{x}^{\mathrm{2n}} \:\mathrm{e}^{\mathrm{x}} }{\mathrm{2n}}\:−\frac{\mathrm{1}}{\mathrm{2n}}\:\int_{\mathrm{0}} ^{\mathrm{x}} \:\mathrm{t}^{\mathrm{2n}} \:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{x}^{\mathrm{2n}} \:\mathrm{e}^{\mathrm{x}} }{\mathrm{2n}}−\frac{\mathrm{1}}{\mathrm{2n}}\mathrm{U}_{\mathrm{n}+\mathrm{1}} \:\Rightarrow\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{x}^{\mathrm{2n}−\mathrm{1}} \:\mathrm{e}^{\mathrm{x}} }{\mathrm{2n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}\left\{\frac{\mathrm{x}^{\mathrm{2n}} \:\mathrm{e}^{\mathrm{x}} }{\mathrm{2n}}−\frac{\mathrm{1}}{\mathrm{2n}}\:\mathrm{U}_{\mathrm{n}+\mathrm{1}} \right\} \\ $$$$=\frac{\mathrm{x}^{\mathrm{2n}−\mathrm{1}} \:\mathrm{e}^{\mathrm{x}} }{\mathrm{2n}−\mathrm{1}}−\frac{\mathrm{x}^{\mathrm{2n}} \:\mathrm{e}^{\mathrm{x}} }{\left(\mathrm{2n}−\mathrm{1}\right)\left(\mathrm{2n}\right)}\:+\frac{\mathrm{1}}{\mathrm{2n}\left(\mathrm{2n}−\mathrm{1}\right)}\:\mathrm{U}_{\mathrm{n}+\mathrm{1}} \\ $$$$…\mathrm{be}\:\mathrm{continued} \\ $$
Answered by maths mind last updated on 12/Jul/20
∫(e^x /(t+x^2 ))dx=f(t) =(1/(2i(√t)))∫{−(e^x /(x+i(√t)))+(e^x /(x−i(√t)))}dx u=x+i(√t)⇒du=dx =−(1/(2i(√t)))∫(e^(u−i(√t)) /u)du+(1/(2i(√t)))∫(e^(u+i(√t)) /u)du =2Re∫(e^(u+i(√t)) /u)du =2Re∫(e^u /u){cos((√t))+isin((√t)))du =2∫(e^u /u)cos((√t))du=2cos((√t))E_i (u) =2cos((√t))E_i (x+i(√t))+c=f(t) we want−f′(1) f′(t)=−2sin((√t))E_i (x+i(√t))+(i/( (√t)))cos((√t)).(e^(x+i(√t)) /(x+i(√t))) −f′(1)=2sin(1)E_i (x+i)−((cos(1)e^x (cos(1)+isin(1)))/(x+i))+c =∫(e^x /((1+x^2 )^2 ))dx
$$\int\frac{{e}^{{x}} }{{t}+{x}^{\mathrm{2}} }{dx}={f}\left({t}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{t}}}\int\left\{−\frac{{e}^{{x}} }{{x}+{i}\sqrt{{t}}}+\frac{{e}^{{x}} }{{x}−{i}\sqrt{{t}}}\right\}{dx} \\ $$$${u}={x}+{i}\sqrt{{t}}\Rightarrow{du}={dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{t}}}\int\frac{{e}^{{u}−{i}\sqrt{{t}}} }{{u}}{du}+\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{t}}}\int\frac{{e}^{{u}+{i}\sqrt{{t}}} }{{u}}{du} \\ $$$$=\mathrm{2}{Re}\int\frac{{e}^{{u}+{i}\sqrt{{t}}} }{{u}}{du} \\ $$$$=\mathrm{2}{Re}\int\frac{{e}^{{u}} }{{u}}\left\{{cos}\left(\sqrt{{t}}\right)+{isin}\left(\sqrt{{t}}\right)\right){du} \\ $$$$=\mathrm{2}\int\frac{{e}^{{u}} }{{u}}{cos}\left(\sqrt{{t}}\right){du}=\mathrm{2}{cos}\left(\sqrt{{t}}\right){E}_{{i}} \left({u}\right) \\ $$$$=\mathrm{2}{cos}\left(\sqrt{{t}}\right){E}_{{i}} \left({x}+{i}\sqrt{{t}}\right)+{c}={f}\left({t}\right) \\ $$$${we}\:{want}−{f}'\left(\mathrm{1}\right) \\ $$$${f}'\left({t}\right)=−\mathrm{2}{sin}\left(\sqrt{{t}}\right){E}_{{i}} \left({x}+{i}\sqrt{{t}}\right)+\frac{{i}}{\:\sqrt{{t}}}{cos}\left(\sqrt{{t}}\right).\frac{{e}^{{x}+{i}\sqrt{{t}}} }{{x}+{i}\sqrt{{t}}} \\ $$$$−{f}'\left(\mathrm{1}\right)=\mathrm{2}{sin}\left(\mathrm{1}\right){E}_{{i}} \left({x}+{i}\right)−\frac{{cos}\left(\mathrm{1}\right){e}^{{x}} \left({cos}\left(\mathrm{1}\right)+{isin}\left(\mathrm{1}\right)\right)}{{x}+{i}}+{c} \\ $$$$=\int\frac{{e}^{{x}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *