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Question Number 176679 by mnjuly1970 last updated on 24/Sep/22
Eeasy integral.... 𝛀 = ∫_(βˆ’βˆ«_0 ^( ∞) e^( βˆ’x^( 2) ) dx) ^( ∫_0 ^( ∞) e^( βˆ’x^( 2) ) dx) sin^( 2) (t).ln^( 3) ( t + (√(1+t^( 2) )))dt βˆ’βˆ’βˆ’m.nβˆ’βˆ’βˆ’
$$ \\ $$$$\:\:\:\:\:\:\:\:{Eeasy}\:\:{integral}…. \\ $$$$\:\:\:\:\:\:\boldsymbol{\Omega}\:=\:\int_{βˆ’\int_{\mathrm{0}} ^{\:\infty} {e}^{\:βˆ’{x}^{\:\mathrm{2}} } {dx}} ^{\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:βˆ’{x}^{\:\mathrm{2}} } {dx}} {sin}^{\:\mathrm{2}} \left({t}\right).{ln}^{\:\mathrm{3}} \left(\:{t}\:+\:\sqrt{\mathrm{1}+{t}^{\:\mathrm{2}} }\right){dt}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:βˆ’βˆ’βˆ’{m}.{n}βˆ’βˆ’βˆ’ \\ $$
Answered by Peace last updated on 24/Sep/22
ln(βˆ’t+(√(1+t^2 )))=βˆ’ln(t+(√(1+t^2 ))) Ξ©=0
$${ln}\left(βˆ’{t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)=βˆ’{ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$$\Omega=\mathrm{0} \\ $$
Answered by mahdipoor last updated on 25/Sep/22
βˆ’f(a)=βˆ’[sin^2 (a).ln^3 (a+(√(a^2 +1)))]= sin^2 (a).[βˆ’ln(a+(√(a^2 +1)))]^3 = sin^2 (a).ln^3 ((1/(a+(√(a^2 +1)))))= sin^2 (a).ln^3 (((βˆ’a+(√(a^2 +1)))/([a+(√(a^2 +1))][βˆ’a+(√(a^2 +1))])))= sin^2 (βˆ’a).ln^3 (βˆ’a+(√((βˆ’a)^2 +1)))=f(βˆ’a) β‡’β‡’ βˆ€a , f(ΞΎ).Ξ΄x+f(βˆ’ΞΎ).Ξ΄x=0 β‡’β‡’ get ∫_0 ^∞ e^(βˆ’x^2 ) dx=m ΞΎ_i =i.Ξ΄x β‡’β‡’ 0=lim_(Ξ΄xβ†’0) Ξ£_(i=0) ^(m/Ξ΄x) f(βˆ’ΞΎ_i ).Ξ΄x+f(ΞΎ_i ).Ξ΄x= lim_(Ξ΄xβ†’0) Ξ£_(βˆ’m/Ξ΄x) ^(m/Ξ΄x) f(ΞΎ_i ).Ξ΄x=∫_( βˆ’m) ^( m) f(x)dx
$$βˆ’{f}\left({a}\right)=βˆ’\left[{sin}^{\mathrm{2}} \left({a}\right).{ln}^{\mathrm{3}} \left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)\right]= \\ $$$${sin}^{\mathrm{2}} \left({a}\right).\left[βˆ’{ln}\left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)\right]^{\mathrm{3}} = \\ $$$${sin}^{\mathrm{2}} \left({a}\right).{ln}^{\mathrm{3}} \left(\frac{\mathrm{1}}{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\right)= \\ $$$${sin}^{\mathrm{2}} \left({a}\right).{ln}^{\mathrm{3}} \left(\frac{βˆ’{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}{\left[{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right]\left[βˆ’{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right]}\right)= \\ $$$${sin}^{\mathrm{2}} \left(βˆ’{a}\right).{ln}^{\mathrm{3}} \left(βˆ’{a}+\sqrt{\left(βˆ’{a}\right)^{\mathrm{2}} +\mathrm{1}}\right)={f}\left(βˆ’{a}\right) \\ $$$$\Rightarrow\Rightarrow \\ $$$$\forall{a}\:,\:\:{f}\left(\xi\right).\delta{x}+{f}\left(βˆ’\xi\right).\delta{x}=\mathrm{0} \\ $$$$\Rightarrow\Rightarrow \\ $$$${get}\:\int_{\mathrm{0}} ^{\infty} {e}^{βˆ’{x}^{\mathrm{2}} } {dx}={m} \\ $$$$\xi_{{i}} ={i}.\delta{x} \\ $$$$\Rightarrow\Rightarrow \\ $$$$\mathrm{0}=\underset{\delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\underset{{i}=\mathrm{0}} {\overset{{m}/\delta{x}} {\sum}}{f}\left(βˆ’\xi_{{i}} \right).\delta{x}+{f}\left(\xi_{{i}} \right).\delta{x}= \\ $$$$\underset{\delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\underset{βˆ’{m}/\delta{x}} {\overset{{m}/\delta{x}} {\sum}}{f}\left(\xi_{{i}} \right).\delta{x}=\int_{\:βˆ’{m}} ^{\:{m}} {f}\left({x}\right){dx} \\ $$$$ \\ $$

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