e-x-1-x-2-dx-

Question Number 44654 by manish00@gmail.com last updated on 02/Oct/18

$$\int\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} }{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=? \\ $$

Commented by maxmathsup by imad last updated on 02/Oct/18

changement x=tanθ give I =∫ (e^(tanθ) /(1+tan^2 θ)) (1+tan^2 θ)dθ = ∫ e^(tanθ) dθ =∫ (Σ_(n=0) ^∞ (((tanθ)^n )/(n!)))dθ =Σ_(n=0) ^∞ (1/(n!)) ∫ tan^n dθ and A_n =∫ tan^n θdθ can be calculated by recurrence ...

$${changement}\:{x}={tan}\theta\:{give}\:{I}\:=\int\:\:\frac{{e}^{{tan}\theta} }{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\:\int\:{e}^{{tan}\theta} {d}\theta\:\:=\int\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left({tan}\theta\right)^{{n}} }{{n}!}\right){d}\theta \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\int\:{tan}^{{n}} {d}\theta\:\:\:{and}\:{A}_{{n}} =\int\:{tan}^{{n}} \theta{d}\theta\:{can}\:{be}\:{calculated}\:{by}\:{recurrence}\:… \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 02/Oct/18

we have e^x =Σ_(n=0) ^∞ (x^n /(n!)) ⇒ ∫ (e^x /(1+x^2 ))dx = ∫ (1/(1+x^2 )) (Σ_(n=0) ^∞ (x^n /(n!)))dx =Σ_(n=0) ^∞ (1/(n!)) ∫ (x^n /(1+x^2 )) dx =Σ_(n=0) ^∞ (A_n /(n!)) with A_n = ∫ (x^n /(1+x^2 ))dx ....

$${we}\:{have}\:{e}^{{x}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}!}\:\Rightarrow\:\int\:\frac{{e}^{{x}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}!}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}!}\:\int\:\frac{{x}^{{n}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{A}_{{n}} }{{n}!}\:\:{with}\:{A}_{{n}} =\:\int\:\:\frac{{x}^{{n}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:…. \\ $$

Commented by maxmathsup by imad last updated on 02/Oct/18