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Question Number 37602 by prof Abdo imad last updated on 15/Jun/18
find the value of  f(a)= ∫_0 ^∞    ((x^2 −1)/((x^4    +a^4 )^2 ))dx  2) calculate   ∫_0 ^∞       ((x^2  −1)/((x^4  +1)^2 ))dx
findthevalueoff(a)=0x21(x4+a4)2dx2)calculate0x21(x4+1)2dx
Commented by prof Abdo imad last updated on 16/Jun/18
1) we have f(a)=∫_(−∞) ^(+∞)   ((x^2 −1)/((x^4  +a^4 )^2 ))dx  let ϕ(z)= ((z^2  −1)/((z^4  +a^4 )^2 ))  ϕ(z)= ((z^2  −1)/({ (z^2 )^2 −(ia^2 )^2 }^2 ))  =((z^2  −1)/((z^2 −ia^2 )^2 (z^2  +ia^2 )^2 ))  =((z^2  −1)/((z −a e^(i(π/4)) )^2 ( z+ae^((iπ)/4) )^2 (z−a e^(−((iπ)/4)) )^2 (z+a e^(−((iπ)/4)) )^2 ))  the poles of ϕ are  +^−  e^(i(π/4))   and +^−  e^(−((iπ)/4))   (doubles)  let suppose a>0  ∫_(−∞) ^(+∞)   ϕ(z)dz =2iπ{ Res(ϕ, ae^((iπ)/4) ) +Res(ϕ,−ae^(−((iπ)/4)) )}  Res(ϕ,ae^(i(π/4)) ) =lim_(z→ae^((iπ)/4) )   (1/((2−1)!)){(z−ae^((iπ)/4) )^2 ϕ(z)}^((1))   =lim_(z→ ae^((iπ)/4) )     {  ((z^2 −1)/((z+a e^((iπ)/4) )^2 (z^2  +ia^2 )^2 ))}^((1))   =lim_(z→ ae^((iπ)/4) )    ((2z(z+ae^((iπ)/4) )(z^2  +ia^2 )^2  −(z^2 −1)(2(z+ae^((iπ)/4) )(z^2  +ia^2 )^2  +4z(z^2  +ia^2 )(z+ae^((iπ)/4) )^2 )/((z +ae^((iπ)/4) )^4 ( z^2  +ia^2 )^4 ))  =lim_(z→a e^((iπ)/4) )   ((2z(z^2  +ia^2 )−(z^2 −1){ 2(z^2  +ia^2 )+4z(z+ae^((iπ)/4) )})/((z +a e^((iπ)/4) )^3 (z^2  +ia^2 )^3 ))  =((2a e^((iπ)/4) (2ia^2 ) −(ia^2 −1){4ia^2  +4ae^((iπ)/4)  (2ae^((iπ)/4) )})/((2ae^(i(π/4)) )^3 (2ia^2 )^3 ))  =((4ia^3  e^((iπ)/4)  −(ia^2 −1){ 4ia^2  +8a^2 i))/(8a^3  i e^((iπ)/4)  (−8ia^6 )))  =(( 4ia^3 e^((iπ)/4)   −12ia^2 (ia^2 −1))/(64 a^9  e^((iπ)/4) ))  =((4ia^(3 ) e^((iπ)/4)  +12a^4   +12ia^2 )/(64a^9  e^((iπ)/4) )) =.....
1)wehavef(a)=+x21(x4+a4)2dxletφ(z)=z21(z4+a4)2φ(z)=z21{(z2)2(ia2)2}2=z21(z2ia2)2(z2+ia2)2=z21(zaeiπ4)2(z+aeiπ4)2(zaeiπ4)2(z+aeiπ4)2thepolesofφare+eiπ4and+eiπ4(doubles)letsupposea>0+φ(z)dz=2iπ{Res(φ,aeiπ4)+Res(φ,aeiπ4)}Res(φ,aeiπ4)=limzaeiπ41(21)!{(zaeiπ4)2φ(z)}(1)=limzaeiπ4{z21(z+aeiπ4)2(z2+ia2)2}(1)=limzaeiπ42z(z+aeiπ4)(z2+ia2)2(z21)(2(z+aeiπ4)(z2+ia2)2+4z(z2+ia2)(z+aeiπ4)2(z+aeiπ4)4(z2+ia2)4=limzaeiπ42z(z2+ia2)(z21){2(z2+ia2)+4z(z+aeiπ4)}(z+aeiπ4)3(z2+ia2)3=2aeiπ4(2ia2)(ia21){4ia2+4aeiπ4(2aeiπ4)}(2aeiπ4)3(2ia2)3=4ia3eiπ4(ia21){4ia2+8a2i)8a3ieiπ4(8ia6)=4ia3eiπ412ia2(ia21)64a9eiπ4=4ia3eiπ4+12a4+12ia264a9eiπ4=..
Commented by behi83417@gmail.com last updated on 16/Jun/18
x=a(√(tgt))⇒dx=((a(1+tg^2 t)dt)/(2(√(tgt))))  I=∫((a^2 tgt−1)/((a^4 tg^2 t+a^4 )^2 )).((a(1+tg^2 t)dt)/(2(√(tgt))))  ⇒I=∫((a^2 tgt−1)/(a^8 (tg^2 t+1)^2 )).((a(1+tg^2 t)dt)/(2(√(tgt))))=  =(1/a^7 )∫((a^2 tgt−1)/(2(√(tgt))(1+tg^2 t)))dt,tgt=u^2   (1+tg^2 t)dt=2udu⇒dt=((2udu)/(1+u^4 ))    ⇒I=(1/a^7 )∫((a^2 u^2 −1)/(2u(1+u^4 ))).((2udu)/(1+u^4 ))=(1/a^7 )∫((a^2 u^2 −1)/((1+u^4 )^2 ))du=  =(1/a^7 )[∫((a^2 u^2 )/((1+u^4 )^2 ))du−∫(du/((1+u^4 )^2 ))]=...  part#1)∫((u^2 du)/((1+u^4 )^2 ))=(u^3 /(4(u^4 +1)))+((√2)/(16))tg^(−1) ((u(√2))/(1−u^2 ))+  +((√2)/(16))ln((u^2 −u(√2)+1)/( (√(1+u^4 ))))+const  part#2)∫(du/((1+u^4 )^2 ))=(u/(4(u^4 +1)))+((3(√2))/(16))tg^(−1) ((u(√2))/(1−u^2 ))−  −((3(√2))/(16))ln((u^2 −u(√2)+1)/( (√(1+u^4 ))))+const  by symplifing:  I=(1/a^5 ).part(#1)−(1/a^7 ).part(#2)  change⇒u→(x/a)  I=(1/a^5 )[((ax^3 )/(4(a^4 +x^4 )))+((√2)/(16))tg^(−1) (((ax(√2))/(a^2 −x^2 )))−  −((3(√2))/(16))ln((x^2 −ax(√2)+a^2 )/( (√(a^4 +x^4 ))))]−(1/a^7 )[((a^3 x)/(4(a^4 +x^4 )))+   +((3(√2))/(16))tg^(−1) ((ax(√2))/(a^2 −x^2 ))−((3(√2))/(16))ln((x^2 −ax(√2)+a^2 )/( (√(a^4 +x^4 ))))]+const  3)I=F(∞)−F(0)=0  4)a=1⇒I=(1/8)[((2x)/(1+x^4 ))+2(√2)tg^(−1) ((x(√2))/(1−x^2 ))−  −(√2)ln((x^2 −x(√2)+1)/( (√(1+x^4 ))))]+const .■
x=atgtdx=a(1+tg2t)dt2tgtI=a2tgt1(a4tg2t+a4)2.a(1+tg2t)dt2tgtI=a2tgt1a8(tg2t+1)2.a(1+tg2t)dt2tgt==1a7a2tgt12tgt(1+tg2t)dt,tgt=u2(1+tg2t)dt=2ududt=2udu1+u4I=1a7a2u212u(1+u4).2udu1+u4=1a7a2u21(1+u4)2du==1a7[a2u2(1+u4)2dudu(1+u4)2]=You can't use 'macro parameter character #' in math mode+216lnu2u2+11+u4+constYou can't use 'macro parameter character #' in math mode3216lnu2u2+11+u4+constbysymplifing:You can't use 'macro parameter character #' in math modechangeuxaI=1a5[ax34(a4+x4)+216tg1(ax2a2x2)3216lnx2ax2+a2a4+x4]1a7[a3x4(a4+x4)++3216tg1ax2a2x23216lnx2ax2+a2a4+x4]+const3)I=F()F(0)=04)a=1I=18[2x1+x4+22tg1x21x22lnx2x2+11+x4]+const.◼
Commented by behi83417@gmail.com last updated on 16/Jun/18
dear prof. abdo! is it ok?
dearprof.abdo!isitok?
Commented by prof Abdo imad last updated on 16/Jun/18
sir Behi you have using a good changement  at the begining and its seems that your method  is correct  and i will giveyou the final answer  after finishing the calculus...
sirBehiyouhaveusingagoodchangementatthebegininganditsseemsthatyourmethodiscorrectandiwillgiveyouthefinalanswerafterfinishingthecalculus

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