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Question Number 73178 by mathmax by abdo last updated on 07/Nov/19
calculate ∫_0 ^∞ ((ln(2+x^2 ))/(x^2 −x+1))dx
calculate∫0∞ln(2+x2)x2−x+1dx
Answered by mind is power last updated on 07/Nov/19
=∫_∞ ^0 ((ln(2+(1/x^2 )))/((1/x^2 )−(1/x)+1)).−(dx/x^2 )=∫_0 ^(+∞) ((ln(2x^2 +1)−ln(x^2 ))/(1−x+x^2 ))dx =∫_0 ^(+∞) ((ln(2x^2 +1))/(1−x+x^2 ))dx−2∫_0 ^(+∞) ((ln(x))/(1−x+x^2 )) ∫_0 ^(+∞) ((ln(x))/(1−x+x^2 ))dx=−∫_0 ^(+∞) ((ln(x))/(1−x+x^2 ))dx⇒=0 =∫_0 ^(+∞) ((ln(2x^2 +1))/(1−x+x^2 ))dx let f(t)=∫_0 ^(+∞) ((ln(tx^2 +1))/(1−x+x^2 ))dx t≥0 f(0)=0 f′(t)=∫_0 ^∞ ((x^2 dx)/((tx^2 +1)(1−x+x^2 ))) f′(t)=∫_0 ^(+∞) ((ax+b)/(tx^2 +1))+((cx+d)/(1−x+x^2 )) b+d=0 (a/t)+c=0⇒a+tc=0 a−b+c=0 dt−a+b=1 b=−d a=−tc −tc+d+c=0 dt+tc−d=1 d=c(t−1) c(t−1)^2 +tc=1 c=(1/(t^2 −t+1)),d=((t−1)/(t^2 −t+1)),a=((−t)/(t^2 −t+1)),b=((1−t)/(t^2 −t+1)) f′(t)=(1/(t^2 −t+1))∫_0 ^(+∞) ((−tx+(1−t))/(tx^2 +1))dx+(1/(t^2 −t+1))∫_0 ^(+∞) ((x+t−1)/(x^2 −x+1)) =(1/(t^2 −t+1))∫_0 ^(+∞) ((−tx)/(tx^2 +1))+((1−t)/(t^2 −t+1))∫_0 ^(+∞) (dx/(tx^2 +1))+(1/(t^2 −t+1))∫_0 ^(+∞) ((x−(1/2))/(x^2 −x+1))dx +((t−(1/2))/(t^2 −t+1))∫_0 ^(+∞) (dx/(x^2 −x+1)) =lim_(y→∞) ((1/2)/(t^2 −t+1))ln(((x^2 −x+1)/(tx^2 +1)))_0 ^y =−((ln(t))/(2(t^2 −t+1))) ∫_0 ^(+∞) (dx/(tx^2 +1))=(1/( (√t))).(π/2) ∫_0 ^(+∞) (dx/(x^2 −x+1))=∫_0 ^(+∞) (dx/((x−(1/2))^2 +(3/4)))=(2/( (√3))).{(π/2)+(π/6)}=((4π)/(3(√3))) ⇒f′(t)=((ln(t))/(2(t^2 −t+1)))+((1−t)/(t^2 −t+1)).(π/2)(√t)+((t−(1/2))/(t^2 −t+1)).((4π)/(3(√3))) ⇒our integral is f(2),f(0)=0 ∫_0 ^(+∞) ((ln(2x^2 +1))/(x^2 −x+1))dx=f(2)=∫_0 ^2 f′(t)dt =∫_0 ^2 ((ln(t)dt)/(2(t^2 −t+1)))+(π/2)∫_0 ^2 (((1−t)(√t))/(t^2 −t+1))dt+((4π)/(3(√3)))∫_0 ^2 ((2t−1)/(2(t^2 −t+1)))dt only ∫_0 ^2 ((ln(t))/(t^2 −t+1))dt is tricky t^2 −t+1=(t−((1−i(√3))/2))(t−((1+i(√3))/2)) (1/(t^2 −t+1))=((−i(√3))/(t−((1−i(√3))/2)))+((i(√3))/(t−((1+i(√3))/2))) ∫_0 ^2 ((ln(t))/(t^2 −t+1))dt=−i(√3)∫_0 ^2 ((ln(t))/(t−((1−i(√3))/2)))dt+i(√3)∫_0 ^2 (dt/(t−((1+i(√3))/2)))=c j=((1−i(√3))/2),f=((1+i(√3))/2) c=i(√3)∫_0 ^2 ((ln(t))/(1−ft))dt−i(√3)∫_0 ^2 ((ln(t))/(1−jt))dt u=ft in firt ,jt=s in 2nd c=i(√3)j∫_0 ^(2f) ((ln(u)+ln(j))/(1−u))du−i(√3)f∫_0 ^(2j) ((ln(s)+ln(f))/(1−s))ds ∫_1 ^(1−z) ((ln(t))/(1−t))dt=Li_2 (z) ⇒∫_0 ^(2f) ((ln(u)+ln(j))/(1−u))du=Li_2 (1−2f)−Li_2 (1)−log(j)log(1−2f) ∫_0 ^(2j) ((ln(s)+ln(f))/(1−s))=Li_2 (1−2j)−Li_2 (1)−log(f)log(1−2j) ⇒∫_0 ^2 ((ln(t))/(t^2 −t+1))dt=i(√3)(((1−i(√3))/2)){Li_2 (−i(√3))−Li_2 (1)+((iπ)/3)Log(−i(√3)} −i(√3)(((1+i(√3))/2)){Li_2 (i(√3))−Li_2 (1)−((iπ)/3)log(i(√3))}
=∫∞0ln(2+1x2)1x2−1x+1.−dxx2=∫0+∞ln(2x2+1)−ln(x2)1−x+x2dx=∫0+∞ln(2x2+1)1−x+x2dx−2∫0+∞ln(x)1−x+x2∫0+∞ln(x)1−x+x2dx=−∫0+∞ln(x)1−x+x2dx⇒=0=∫0+∞ln(2x2+1)1−x+x2dxletf(t)=∫0+∞ln(tx2+1)1−x+x2dxt⩾0f(0)=0f′(t)=∫0∞x2dx(tx2+1)(1−x+x2)f′(t)=∫0+∞ax+btx2+1+cx+d1−x+x2b+d=0at+c=0⇒a+tc=0a−b+c=0dt−a+b=1b=−da=−tc−tc+d+c=0dt+tc−d=1d=c(t−1)c(t−1)2+tc=1c=1t2−t+1,d=t−1t2−t+1,a=−tt2−t+1,b=1−tt2−t+1f′(t)=1t2−t+1∫0+∞−tx+(1−t)tx2+1dx+1t2−t+1∫0+∞x+t−1x2−x+1=1t2−t+1∫0+∞−txtx2+1+1−tt2−t+1∫0+∞dxtx2+1+1t2−t+1∫0+∞x−12x2−x+1dx+t−12t2−t+1∫0+∞dxx2−x+1=limy→∞12t2−t+1ln(x2−x+1tx2+1)0y=−ln(t)2(t2−t+1)∫0+∞dxtx2+1=1t.π2∫0+∞dxx2−x+1=∫0+∞dx(x−12)2+34=23.{π2+π6}=4π33⇒f′(t)=ln(t)2(t2−t+1)+1−tt2−t+1.π2t+t−12t2−t+1.4π33⇒ourintegralisf(2),f(0)=0∫0+∞ln(2x2+1)x2−x+1dx=f(2)=∫02f′(t)dt=∫02ln(t)dt2(t2−t+1)+π2∫02(1−t)tt2−t+1dt+4π33∫022t−12(t2−t+1)dtonly∫02ln(t)t2−t+1dtistrickyt2−t+1=(t−1−i32)(t−1+i32)1t2−t+1=−i3t−1−i32+i3t−1+i32∫02ln(t)t2−t+1dt=−i3∫02ln(t)t−1−i32dt+i3∫02dtt−1+i32=cj=1−i32,f=1+i32c=i3∫02ln(t)1−ftdt−i3∫02ln(t)1−jtdtu=ftinfirt,jt=sin2ndc=i3j∫02fln(u)+ln(j)1−udu−i3f∫02jln(s)+ln(f)1−sds∫11−zln(t)1−tdt=Li2(z)⇒∫02fln(u)+ln(j)1−udu=Li2(1−2f)−Li2(1)−log(j)log(1−2f)∫02jln(s)+ln(f)1−s=Li2(1−2j)−Li2(1)−log(f)log(1−2j)⇒∫02ln(t)t2−t+1dt=i3(1−i32){Li2(−i3)−Li2(1)+iπ3Log(−i3}−i3(1+i32){Li2(i3)−Li2(1)−iπ3log(i3)}
Commented by mathmax by abdo last updated on 07/Nov/19
thankx sir.
thankxsir.
Commented by mind is power last updated on 07/Nov/19
y′re welcom sir
y′rewelcomsir
Commented by aliesam last updated on 15/Dec/19
I didn′t get the first step where you factor −(dx/x^2 ) i mean the ln part
Ididn′tgetthefirststepwhereyoufactor−dxx2imeanthelnpart
Commented by aliesam last updated on 15/Dec/19
I mean ∫_0 ^∞ ((ln(2+x^2 ))/(x^2 −x+1))=∫_∞ ^0 ((ln(2+(1/x^2 )))/((1/x^2 )−(1/x)+1)) . ((−dx)/x^2 ) ??
Imean∫0∞ln(2+x2)x2−x+1=∫∞0ln(2+1x2)1x2−1x+1.−dxx2??

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