answer to 25955.we introduce the parametric function
F(t) =∫_0 ^∞ ln(1+(1+x^2_ )t)(1+x^2 )^(−1) dx after verifying that
F is derivable on[0.∝[ we find ∂F/∂t= ∫_0 ^∞ ( (1+(1+x^2 )t)^(−1) dx
∂F/∂t=1/2 ∫_R (tx^2 +t+1)^(−1) dx we put f(z) =(tz^2 +z+1)^(−1)
let find the poles of f..tz^2 +z+1=0 <−> z=+−i((t+1)t^(−1) )^(1/2)
and the poles are z_0 =i((t+1)t^(−1) )^(1/2) and z_1 =−i((t+1)t^(−1) )^(1/2)
and f(t) =(t(t−z_0 )(t−z_1 ))^(−1) by residus theorem
∫_R f(z)dz =2iπ R(f.z_0 ) =2iπ (t(z_0 −z_1 ))^(−1)
=π t^(−1/2) (t+1)^(−1/2 ) −>∂F/∂t =π 2^(−1) t^(−1/2) (1+t)^(−1/2)
−>F(t) =π 2^(−1 ) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx +α
α=F(0)=0 and F(t) =π2^(−1) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx
and by the changement x^(1/2) =u we find
F(t) = π ln( t^(1/2) +(1+t)^(1/2) ) so ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx=F(1)=πln(1+2^(1/2) )
answer to 25824 we have a^(−x^2 ) = e^(−x^2_ ln(a)) so for a>1
ln(a)=( (ln(a))^(1/2) )^2 >>>>∫_R a^(−^ x^2 ) = ∫_R e^(−(x (ln(a)^(1/2) )^2 ) dx
and with the changement t=x (ln(a)^(1/2) >>>>x=t ( ln(a))^(−1/2)
we have ∫_R a^(−x^2 ) dx = π^(1/2) (ln(a))^(−1/2) ...if 0<a<1 ln(a)<0
and the integrale is divergente...
find the value of∫_0 ^∞ artan(2x)(1+x^2 )^(−1_ ) the key of slution put F(t)=∫_0 ^∞ artan(xt)(1+x^2 )^(−1) find ∂F/∂t first then F(t) and take t=2 you will of find find the value of integral..