Question Number 37352 by math khazana by abdo last updated on 12/Jun/18
$${let}\:\:{f}\left({z}\right)\:=\:{e}^{−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }} \:\: \\ $$$$\left.\mathrm{1}\right)\:{give}\:{f}\left({z}\right)\:{at}\:{form}\:{of}\:{serie} \\ $$$$\left.\mathrm{2}\right)\:{give}\:\:\int_{\mathrm{1}} ^{\mathrm{2}} {f}\left({z}\right){dz}\:\:\:{at}\:{form}\:{of}\:{serie}\:. \\ $$
Commented by prof Abdo imad last updated on 13/Jun/18
![we have e^(−u) =Σ_(n=0) ^∞ (((−1)^n u^n )/(n!)) for u∈C e^(−(1/z^2 )) = Σ_(n=0) ^∞ (((−1)^n )/(n! z^(2n) )) ⇒ f(z)=Σ_(n=0) ^∞ (((−1)^n )/(n!))(1/z^(2n) ) 2) ∫_1 ^2 f(z)dz = ∫_1 ^2 (Σ_(n=0) ^∞ (((−1)^n )/(n!)) (1/z^(2n) ))dz =Σ_(n=0) ^∞ (((−1)^n )/(n!)) ∫_1 ^2 z^(−2n) dz but ∫_1 ^2 z^(−2n) dz = [(1/(1−2n))z^(1−2n) ]_1 ^2 =−(1/(2n−1)){ (1/2^(2n−1) ) −1} ⇒ ∫_1 ^2 f(z)dz =Σ_(n=0) ^∞ (((−1)^n )/((2n−1)n!)){1−(1/2^(2n−1) )} .](https://www.tinkutara.com/question/Q37430.png)
$${we}\:{have}\:\:{e}^{−{u}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{u}^{{n}} }{{n}!}\:{for}\:{u}\in{C} \\ $$$${e}^{−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }} =\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\:{z}^{\mathrm{2}{n}} }\:\Rightarrow\:{f}\left({z}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\frac{\mathrm{1}}{{z}^{\mathrm{2}{n}} } \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{1}} ^{\mathrm{2}} {f}\left({z}\right){dz}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\frac{\mathrm{1}}{{z}^{\mathrm{2}{n}} }\right){dz} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:{z}^{−\mathrm{2}{n}} {dz}\:{but} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:{z}^{−\mathrm{2}{n}} {dz}\:=\:\left[\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}{n}}{z}^{\mathrm{1}−\mathrm{2}{n}} \right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\left\{\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\:−\mathrm{1}\right\}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:{f}\left({z}\right){dz}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right){n}!}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\right\}\:. \\ $$$$ \\ $$