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Question Number 125708 by Dwaipayan Shikari last updated on 13/Dec/20

∫_0 ^∞ ((1−tanhx)/( ((tanhx))^(1/5) ))dx

01tanhxtanhx5dx

Answered by mindispower last updated on 13/Dec/20

e^x =t  =∫_1 ^∞ ((1−((t−(1/t))/(t+(1/t))))/( (((t^2 −1)/(t^2 +1)))^(1/5) )).(1/t)dt  =∫_1 ^∞ ((2/(t^2 +1))/( (((t^2 −1)/(t^2 +1)))^(1/5) )).(dt/t)=2∫_1 ^∞ (dt/(t(t^2 +1)^(4/5) (t^2 −1)^(1/5) ))  =∫_1 ^∞ ((d(t^2 ))/(t^2 (t^2 +1)^(4/5) (t^2 −1)^(1/5) ))   =∫_1 ^∞ (dt/(t(t+1)^(4/5) (t−1)^(1/5) ))  =∫_0 ^1 ((−d((1/t)))/((1/t)(1+t)^(4/5) (1−t)^(1/5) .(1/t)))=∫_0 ^1 (1+t)^(−(4/5)) (1−t)^(−(1/5)) dt  =∫_0 ^1 t^(1−1) (1−t)^((9/5)−1−1) (1−(−1)t)^(−(4/5)) dt  =β(1,(4/5))     _2 F_1 ((4/5),1;(9/5);−1)  ∫_0 ^∞ ((1−th(x))/( ((th(x)))^(1/5) ))dx=β(1,(4/5))._2 F_1 ((4/5),1;(9/5);1)⋍0.90323

ex=t=11t1tt+1tt21t2+15.1tdt=12t2+1t21t2+15.dtt=21dtt(t2+1)45(t21)15=1d(t2)t2(t2+1)45(t21)15=1dtt(t+1)45(t1)15=01d(1t)1t(1+t)45(1t)15.1t=01(1+t)45(1t)15dt=01t11(1t)9511(1(1)t)45dt=β(1,45)2F1(45,1;95;1)01th(x)th(x)5dx=β(1,45).2F1(45,1;95;1)0.90323

Commented by Dwaipayan Shikari last updated on 13/Dec/20

Great sir! great approach .  My thinking  ∫_0 ^∞ ((1−tanhx)/( ((tanhx))^(1/5) ))dx= ∫_0 ^∞ ((1−tanh^2 x)/((1+tanhx)((tanhx))^(1/5) ))dx    tanhx=t  =∫_0 ^1 (dt/((1+t)(t)^(1/5) ))=5∫_0 ^1 (u^3 /(1+u^5 ))dt ....      t=u^5    We can decompose it

Greatsir!greatapproach.Mythinking01tanhxtanhx5dx=01tanh2x(1+tanhx)tanhx5dxtanhx=t=01dt(1+t)t5=501u31+u5dt....t=u5Wecandecomposeit

Commented by mindispower last updated on 13/Dec/20

yes nice  nice

yesnicenice

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