Question Number 101828 by floor(10²Eta[1]) last updated on 05/Jul/20

∫_0 ^∞ ((e^(πx) −e^x )/(x(e^(πx) +1)(e^x +1)))dx

Answered by maths mind last updated on 05/Jul/20

=∫_0 ^(+∞) (1/x){(1/((e^x +1)))−(1/(e^(πx) +1))}dx  let f bee lik t≥0 well defind C_∞   f(t)=∫_0 ^(+∞) (1/x){(1/(e^x +1))−(1/(e^(πx) +1))}e^(−xt) dx  f′(t)=∫_0 ^(+∞) −(e^(−xt) /(e^x +1))dx+∫_0 ^(+∞) (e^(−xt) /(e^(πx) +1))dx  =−∫_0 ^∞ e^(−x(t+1)) Σ_(k≥0) (−e^(−x) )^k dx+∫_0 ^(+∞) e^(−(t+π)) Σ(−e^(−πx) )^k dx  =Σ_(k≥0) (−1)^(k+1) .(1/(k+t+1))+Σ_(k≥0) (((−1)^k )/(πk+t+π))  =Σ_(k≥0) ((((−1)^k )/(k+t+π))−(((−1)^k )/(k+t+1)))  =Σ_(k≥0) ((1/(2k+t+π))−(1/(2k+t+1)))+Σ_(k≥0) ((1/(2k+2+t))−(1/(2k+t+π+1)))  (1/(t+π))−(1/(t+1))+(1/(2+t))−(1/(t+π+1))+Σ_(k≥1) (−(1/(2k))+(1/(2k+t+π))+(1/(2k))−(1/(2k+t+1)))  +Σ_(k≥1) ((1/(2k+2+t))−(1/(2k))+(1/(2k))−(1/(2k+t+π+1)))  we have H(z)=Σ_(k≥1) ((1/k)−(1/(k+z)))^� ∀z∈C−{Z_− }  we get f′(t)=(1/(t+π))−(1/(t+1))+(1/(t+2))−(1/(t+π+1))−(1/2)H(((t+π)/2))+(1/2)H(((t+1)/2))  −(1/2)H(((t+2)/2))+(1/2)H(((t+π+1)/2))  H(z)=Ψ(z+1)+γ  we get  f′(t)=(1/(t+π))−(1/(t+1))+(1/(t+2))−(1/(t+π+1))+(1/2)(Ψ(((t+3)/2))−Ψ(((t+π+2)/2))+Ψ(((t+π+1)/2))−Ψ(((t+2)/2)))  f(t)=ln(((t+π)/(t+π+1)))+ln(((t+2)/(t+1)))+log(((Γ(((t+3)/2)))/(Γ(((t+2)/2)))))+log(((Γ(((t+π+1)/2)))/(Γ(((t+π+2)/2)))))+c  wr used∫Ψ(z)dz=log(Γ(z))+c  lim_(t→∞) f(t)=0  we need too finde thislim_(x→∞) ((Γ(x+(1/2)))/(Γ(x)))  Γ(z+a)∼(√(2πz)).z^(z+a−(1/2)) e^(−z)   f(t)→0  f(0)=∫_0 ^(+∞) ((e^(πx) −e^x )/(x(e^x +1)(e^(πx) +1)))dx=ln((π/(π+1)))+ln(2)+log(((Γ((3/2))Γ(((π+1)/2)))/(Γ(1)Γ(((π+2)/2)))))

Commented byfloor(10²Eta[1]) last updated on 05/Jul/20

someone help me with the solution:  I(t)=∫_0 ^∞ ((e^(tx) −e^x )/(x(e^(tx) +1)(e^x +1)))dx  now i′m gonna derivate and integrate again:  I′(t)=∫_0 ^∞ (∂/∂t)(((e^(tx) −e^x )/(x(e^(tx) +1)(e^x +1))))dx  =∫_0 ^∞ (e^(tx) /((e^(tx) +1)^2 ))dx, u=e^(tx) +1⇒(du/t)=e^(tx) dx  =(1/t)∫_2 ^∞ (du/u^2 )=(1/(2t))  ⇒I′(t)=(1/(2t))⇒I(t)=∫(1/(2t))=(1/2)ln∣t∣+C  I(1)=0=C⇒I(t)=(1/2)ln∣t∣  ⇒I(π)=ln((√π))

Answered by mathmax by abdo last updated on 05/Jul/20

I =∫_0 ^∞  ((e^(πx) −e^x )/(x(e^(πx)  +1)(e^x  +1)))dx ⇒ I =∫_0 ^∞  ((e^(πx) +1−(1+e^x ))/(x(e^(πx)  +1)(e^x  +1)))dx  I =∫_0 ^∞   (dx/(x(e^x  +1))) −∫_0 ^∞  (dx/(x(e^(πx)  +1))) =I_1  −I_2   we have I_1 =∫_0 ^∞  (e^(−x) /(x(1+e^(−x) )))dx  I_2 =∫_0 ^∞  (e^(−πx) /(x(1+e^(−πx) )))dx =_(πx =t)    ∫_0 ^∞   (e^(−t) /((t/π)(1+e^(−t) )))((dt/π)) =∫_0 ^∞   (e^(−t) /(t(1+e^(−t) )))dt =I_1  ⇒  I =0

Commented bymaths mind last updated on 05/Jul/20

 I_(1,) I_2  didnt  cv

Commented bymathmax by abdo last updated on 05/Jul/20

but I_1 =I_2  ...!i dont see the convergence...