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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 bymath 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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