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Question Number 102985 by DGmichael last updated on 12/Jul/20

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