Question Number 114330 by mnjuly1970 last updated on 18/Sep/20

Commented bymnjuly1970 last updated on 18/Sep/20

easy question↑↑↑

Answered by Dwaipayan Shikari last updated on 18/Sep/20

λ_1 =∫_0 ^∞ ((log(1+x^2 ))/(1+x^2 ))dx  ∫_0 ^(π/2) ((log(1+tan^2 θ))/(sec^2 θ)).sec^2 θdθ  −2∫_0 ^(π/2) log(cosθ)dθ=πlog(2)  λ_2 =∫_0 ^(π/4) ((log(1+tanθ))/(sec^2 θ)).sec^2 θdθ  =∫_0 ^(π/4) log(cosx+sinx)−log(cosx)=∫_0 ^(π/4) (1/2)log2+log(cos((π/4)−x))−log(cosx)=I  =(π/8)log(2)+0=(π/8)log(2)  (λ_1 /λ_2 )=((πlog(2))/((π/8)log(2)))=8

Commented bymnjuly1970 last updated on 18/Sep/20

thank you sir .very nice.

Answered by mathmax by abdo last updated on 18/Sep/20

we have λ_1 =∫_0 ^∞  ((ln(1+x^2 ))/(1+x^2 ))dx =_(x=tanθ)   ∫_0 ^(π/2)  ((ln(1+tan^2 θ))/(1+tan^2 θ))(1+tan^2 θ)dθ  =∫_0 ^(π/2) ln((1/(cos^2 θ)))dθ =−2∫_0 ^(π/2)  ln(cosθ)dθ =−2(−(π/2)ln(2))=πln(2)  λ_2 =∫_0 ^1  ((ln(1+x))/(1+x^2 ))dx =_(x=tant)    ∫_0 ^(π/4)  ((ln(1+tant))/(1+tan^2 t))(1+tan^2 t)dt  =∫_0 ^(π/4)  ln(1+tant)dt =_(t=(π/4)−u)   ∫_0 ^(π/4) ln(1+tan((π/4)−u) du  =∫_0 ^(π/4)  ln(1+((1−tanu)/(1+tanu)))du =∫_0 ^(π/4) ln(  (2/(1+tanu)))du  =(π/4)ln(2) −∫_0 ^(π/4) ln(1+tanu)du =(π/4)ln(2)−λ_2  ⇒  2λ_2 =(π/4)ln(2) ⇒λ_2 =(π/8)ln(2) ⇒(λ_1 /λ_2 ) =((πln(2))/((π/8)ln(2))) ⇒★(λ_1 /λ_2 ) =8 ★

Commented bymnjuly1970 last updated on 18/Sep/20

thank you master (ostad) in  our language.very nice.grateful     teful

Commented bymathmax by abdo last updated on 18/Sep/20

you are welcome sir