I have in trouble for evaluating following integral ∫0∞(1+x4−x2)dx=Γ2(14)6π

Question

I have in trouble for evaluating following integral  ∫0∞(1+x4−x2)dx=Γ2(14)6π

star233 · Accepted Answer

Let x=tan⁡θ. Note that, x4=tan2⁡θ and dx=sec2⁡θdθ2tan⁡θ Like this, I=∫0∞(1+x4−x2)dx=∫0π/2(sec⁡θ+tan⁡θ)sec2⁡θdθ2tan⁡θ=12∫0π/2cos1/2⁡θ(1−sin⁡θ)dθsin1/2⁡θcos3⁡θ =12∫0π/2cos−5/2⁡θsin−1/2⁡θdθ−12∫0π/2cos−5/2⁡θsin1/2⁡dθ=I1+I2 Note that, I1=12B(−3/4,1/4), because (2m−1=−5/2⇒m=−3/4 and 2n−1=−1/2⇒n=1/4) But, B(m,n)=12Γ(m)Γ(n)Γ(m+n). Thus, I1=14Γ(−3/4)Γ(1/4)Γ(−1/2)=14⋅(−4)3Γ(1/4)⋅Γ(1/4)(−2)π=Γ(1/4)26π If x→0,Γ(x)→∞, so that I2=0. Furthermore, if x&amp;lt;0, x≠−1,−2,...,define Γ(x)=Γ(x+1)/x

user_27qwe · Accepted Answer

Another approach : Set x4=1t−1⇒x=(1−tt)14⇒dx=−14t−54(1−t)−34dt Then the integral turms out to be ∫0∞(1+x4−x2)dx=14∫01(t−74(1−t)−34−t−74(1−t)−14)dt=14[B(−34,14)−B(−34,34)]=14[B(−34,14)−0]=14⋅Γ(−34)Γ(14)Γ(−12)=Γ2(14)6π

Vasquez · Accepted Answer

Let&#039;s apply the substitution x4+1−x2=t: x2=1−t2t, dx=−28t−5/4(t+1)(1−t)−1/2dt∫0∞(x4+1−x2)dx=28(∫01t1/4(1−t)−1/2dt+∫01t−3/4(1−t)−1/2dt=28(B(54,12)+B(14,12))=Γ2(14)6π

I have in trouble for evaluating following integral \int_0^\infty(\sqrt{1+x^4}-x^2)dx=\frac{\Gamma^2(\frac{1}{4})}{6\sqrt{\pi}}

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