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Question Number 33352 by caravan msup abdo. last updated on 15/Apr/18
let give S(x)=Σ_(n≥0) (((−1)^n )/( (√(x+n)))) ,x>0 1)study the contnuity ,derivsbility,limits at 0^+ and +∞ 2) we give ∫_0 ^∞ e^(−t^2 ) dt =((√π)/2) .prove that ∀ x>0 S(x)=(1/( (√π))) ∫_0 ^∞ (e^(−tx) /( (√t)(1+e^(−t) )))dt .
$${let}\:{give}\:{S}\left({x}\right)=\sum_{{n}\geqslant\mathrm{0}} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\:\sqrt{{x}+{n}}}\:,{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right){study}\:{the}\:{contnuity}\:,{derivsbility},{limits} \\ $$$${at}\:\mathrm{0}^{+} \:{and}\:+\infty \\ $$$$\left.\mathrm{2}\right)\:{we}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:.{prove}\:{that} \\ $$$$\forall\:{x}>\mathrm{0}\:\:{S}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{tx}} }{\:\sqrt{{t}}\left(\mathrm{1}+{e}^{−{t}} \right)}{dt}\:. \\ $$
Commented by math khazana by abdo last updated on 20/Apr/18
1) let put f_n (x) = (((−1)^n )/( (√(x+n)))) the sequence(f_n ) are continue nd the serie Σ f_n (x) is convergent( if we take ϕ(t)= (1/( (√(x+t)))) .ϕ is decreszing so Σ f_n is alternating) its sum S(x) is continue for x>0 also the (f_n ) are derivables and f_n ^′ (x) = (−1)^n ((−1)/(2(x+n)(√(x+n)))) Σ^ f_n ^′ (x) is convergent as a alternating serie so S is derivable and S^′ (x) = Σ_(n=0) ^∞ (((−1)^(n+1) )/(2(x+n)(√(x+n)))) . 2) we have ∫_0 ^∞ (e^(−tx) /( (√t)(1+e^(−t) )))dt = ∫_0 ^∞ (e^(−tx) /( (√t))) (Σ_(n=0) ^∞ (−1)^n e^(−nt) )dt = Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ (e^(−(n+x)t) /( (√t))) dt (ch.(√t)=u) =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ (e^(−(n+x)u^2 ) /u) 2udu = 2 Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−(n+x)u^2 ) du =_((√(n+x))u =t) 2Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−t^2 ) (dt/( (√(n+x)))) = 2 .((√π)/2) Σ_(n=0) ^∞ (((−1)^n )/( (√(n+x)))) =(√π) S(x) ⇒ S(x) =(1/( (√π))) ∫_0 ^∞ (e^(−tx) /( (√t)( 1+e^(−t) )))dt .
$$\left.\mathrm{1}\right)\:{let}\:{put}\:{f}_{{n}} \left({x}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\:\sqrt{{x}+{n}}}\:\:\:{the}\:{sequence}\left({f}_{{n}} \right)\:{are}\:{continue} \\ $$$${nd}\:{the}\:{serie}\:\Sigma\:{f}_{{n}} \left({x}\right)\:{is}\:{convergent}\left(\:{if}\:{we}\:{take}\right. \\ $$$$\left.\varphi\left({t}\right)=\:\frac{\mathrm{1}}{\:\sqrt{{x}+{t}}}\:.\varphi\:{is}\:{decreszing}\:\:{so}\:\Sigma\:{f}_{{n}} \:{is}\:{alternating}\right) \\ $$$${its}\:{sum}\:{S}\left({x}\right)\:{is}\:{continue}\:{for}\:{x}>\mathrm{0}\:{also}\:{the}\:\left({f}_{{n}} \right)\:{are} \\ $$$${derivables}\:\:{and}\:{f}_{{n}} ^{'} \left({x}\right)\:=\:\left(−\mathrm{1}\right)^{{n}} \:\:\:\frac{−\mathrm{1}}{\mathrm{2}\left({x}+{n}\right)\sqrt{{x}+{n}}} \\ $$$$\sum^{} \:\:\:{f}_{{n}} ^{'} \left({x}\right)\:{is}\:{convergent}\:{as}\:{a}\:{alternating}\:{serie}\:{so} \\ $$$${S}\:{is}\:{derivable}\:{and}\:{S}^{'} \left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}\left({x}+{n}\right)\sqrt{{x}+{n}}}\:. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{e}^{−{tx}} }{\:\sqrt{{t}}\left(\mathrm{1}+{e}^{−{t}} \right)}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{tx}} }{\:\sqrt{{t}}}\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−{nt}} \right){dt} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\left({n}+{x}\right){t}} }{\:\sqrt{{t}}}\:{dt}\:\left({ch}.\sqrt{{t}}={u}\right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−\left({n}+{x}\right){u}^{\mathrm{2}} } }{{u}}\:\mathrm{2}{udu} \\ $$$$=\:\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({n}+{x}\right){u}^{\mathrm{2}} } {du} \\ $$$$=_{\sqrt{{n}+{x}}{u}\:={t}} \:\:\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\:\frac{{dt}}{\:\sqrt{{n}+{x}}} \\ $$$$=\:\mathrm{2}\:.\frac{\sqrt{\pi}}{\mathrm{2}}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\:\sqrt{{n}+{x}}}\:\:=\sqrt{\pi}\:\:{S}\left({x}\right)\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{tx}} }{\:\sqrt{{t}}\left(\:\mathrm{1}+{e}^{−{t}} \right)}{dt}\:. \\ $$

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