Integration Challenge 3! [Courtesy: Hasan Kassim]

∫0∞xa1+xnln⁡(1+xnxm) dx = πncsc⁡(π(a+1)n)(mπncot⁡(π(a+1)n)−(ψ(n−a−1n)+γ))\large \displaystyle \int_{0}^{\infty}{\frac{{x}^{a}}{1+{x}^{n}} \ln\left(\frac{1+{x}^{n}}{{x}^{m}}\right) \ dx} \ = \ \large \frac{\pi}{n} \csc\left(\frac{\pi(a+1)}{n}\right) \left(\frac{m\pi}{n}\cot\left(\frac{\pi(a+1)}{n}\right) - \left(\psi\left(\frac{n-a-1}{n}\right) + \gamma\right)\right)

Prove the identity above.

Note by Kartik Sharma
5 years, 9 months ago

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Comments

Consider I(z)=∫0∞xa(xn+1)zdx\displaystyle\text{I}(z) = \int_{0}^{\infty}\dfrac{x^a}{(x^n+1)^z} \mathrm{d}x

Putting xn+1=1tx^n+1=\dfrac{1}{t}, we have,

I(z)=1n∫01tz−(a+1n−1)(1−t)(a+1n−1)dt\displaystyle\text{I}(z) = \dfrac{1}{n} \int_{0}^{1} t^{z-\left(\frac{a+1}{n}-1\right)}(1-t)^{\left(\frac{a+1}{n}-1\right)} \mathrm{d}t

=1nB⁡(a+1n,z−a+1n)\displaystyle = \dfrac{1}{n} \operatorname{B}\left(\dfrac{a+1}{n}, z-\dfrac{a+1}{n}\right)

=1nΓ(a+1n)Γ(z−a+1n)Γ(z)\displaystyle=\dfrac{1}{n} \dfrac{\Gamma \left(\dfrac{a+1}{n}\right)\Gamma\left(z-\dfrac{a+1}{n}\right)}{\Gamma(z)}

  ⟹  I′(z)=1nΓ(a+1n)Γ′(z−a+1n)Γ(z)−Γ′(z)Γ(a+1n)Γ(z−a+1n)(Γ(z))2\displaystyle\implies \text{I} '(z)=\dfrac{1}{n} \dfrac{\Gamma \left(\dfrac{a+1}{n}\right)\Gamma ' \left(z-\dfrac{a+1}{n}\right)\Gamma(z)-\Gamma '(z)\Gamma\left(\dfrac{a+1}{n}\right)\Gamma\left(z-\dfrac{a+1}{n}\right)}{(\Gamma (z))^2}

  ⟹  I′(1)=1nΓ(a+1n)Γ′(1−a+1n)Γ(1)−Γ′(1)Γ(a+1n)Γ(1−a+1n)(Γ(1))2\displaystyle\implies \text{I} '(1)=\dfrac{1}{n} \dfrac{\Gamma \left(\dfrac{a+1}{n}\right)\Gamma ' \left(1-\dfrac{a+1}{n}\right)\Gamma(1)-\Gamma '(1)\Gamma\left(\dfrac{a+1}{n}\right)\Gamma\left(1-\dfrac{a+1}{n}\right)}{(\Gamma (1))^2}

Now, since Γ′(z)=ψ(z)⋅Γ(z)\displaystyle\Gamma '(z)=\psi(z)\cdot\Gamma(z) and Γ′(1)=−γ\Gamma '(1)=-\gamma and using Euler's reflection formula, we have,

I′(1)=πncsc⁡(π(a+1)n(ψ(1−a+1n)+γ))\displaystyle\text{I}'(1)=\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right)

Also, from the original integral,

I′(1)=−∫0∞xaxn+1ln⁡(xn+1) dx\displaystyle\text{I}'(1)=-\int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln (x^n+1) \ \mathrm{d}x

∴∫0∞xaxn+1ln⁡(xn+1) dx=−πncsc⁡(π(a+1)n(ψ(1−a+1n)+γ)) (1)\displaystyle\therefore \int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln (x^n+1) \ \mathrm{d}x = -\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right) \ (1)

Next, let J(a)=∫0∞xa(xn+1)dx\displaystyle\text{J}(a)= \int_{0}^{\infty}\dfrac{x^a}{(x^n+1)}\mathrm{d}x

Note that J(a)=I(1)\text{J}(a)=\text{I}(1)

=1nΓ(a+1n)Γ(1−a+1n)Γ(1)\displaystyle=\dfrac{1}{n} \dfrac{\Gamma \left(\dfrac{a+1}{n}\right)\Gamma\left(1-\dfrac{a+1}{n}\right)}{\Gamma(1)}

=πncsc⁡(π(a+1)n)\displaystyle=\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\right)

Therefore,

J′(1)=∫0∞xa1+xnln⁡x dx=−π2n2csc⁡(π(a+1)n)cot⁡(π(a+1)n) (2)\displaystyle\text{J}'(1)=\int_{0}^{\infty}\dfrac{x^a}{1+x^n}\ln x \ \mathrm{d}x = -\dfrac{{\pi}^{2}}{n^2}\csc\left(\dfrac{\pi(a+1)}{n}\right)\cot\left(\dfrac{\pi(a+1)}{n}\right) \ (2)

Operating (1)−m×(2)(1)-m\times(2), we have,

∫0∞xaxn+1ln⁡(xn+1xm)dx=πncsc⁡(π(a+1)n)[mπncot⁡(π(a+1)n)−(ψ(1−a+1n)+γ)]\displaystyle\int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln \left(\dfrac{x^n+1}{x^m}\right)\mathrm{d}x=\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\right)\left[\dfrac{m\pi}{n}\cot\left(\dfrac{\pi(a+1)}{n}\right)-\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right]

Q.E.D.

Ishan Singh - 5 years, 9 months ago

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@Kartik Sharma Can you please re-check whether R.H.S. contains +γ+\gamma or −γ-\gamma in the bracket?

Ishan Singh - 5 years, 9 months ago

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Edited! Great! Congratulations once again! (You can post the proof of the first challenge even though I've done already, a new proof will be accepted; just in case I thought you must be interested).

Kartik Sharma - 5 years, 9 months ago

@Hasan Kassim

Kartik Sharma - 5 years, 9 months ago

I'm not so good at high level integrals. But I'm writing all the proofs provided by u and trying to use them. Thanks a lot man. Keep on posting proofs!

Aditya Kumar - 5 years, 9 months ago

so LoveLy As always

Aman Rajput - 5 years, 9 months ago
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