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Integration Challenge 3! [Courtesy: Hasan Kassim]

\[\large \displaystyle \int_{0}^{\infty}{\frac{{x}^{a}}{1+{x}^{n}} \ln\left(\frac{1+{x}^{n}}{{x}^{m}}\right) \ dx} \ = \ \large \frac{\pi}{n} \csc\left(\frac{\pi(a+1)}{n}\right) \left(\frac{m\pi}{n}\cot\left(\frac{\pi(a+1)}{n}\right) - \left(\psi\left(\frac{n-a-1}{n}\right) + \gamma\right)\right)\]

Prove the identity above.

Note by Kartik Sharma
2 years, 5 months ago

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

Putting \(x^n+1=\dfrac{1}{t}\), we have,

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

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

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

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

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

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

\(\displaystyle\text{I}'(1)=\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right)\)

Also, from the original integral,

\(\displaystyle\text{I}'(1)=-\int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln (x^n+1) \ \mathrm{d}x\)

\(\displaystyle\therefore \int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln (x^n+1) \ \mathrm{d}x = -\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right) \ (1)\)

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

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

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

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

Therefore,

\(\displaystyle\text{J}'(1)=\int_{0}^{\infty}\dfrac{x^a}{1+x^n}\ln x \ \mathrm{d}x = -\dfrac{{\pi}^{2}}{n^2}\csc\left(\dfrac{\pi(a+1)}{n}\right)\cot\left(\dfrac{\pi(a+1)}{n}\right) \ (2)\)

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

\(\displaystyle\int_{0}^{\infty}\dfrac{x^a}{x^n+1}\ln \left(\dfrac{x^n+1}{x^m}\right)\mathrm{d}x=\dfrac{\pi}{n}\csc\left(\dfrac{\pi(a+1)}{n}\right)\left[\dfrac{m\pi}{n}\cot\left(\dfrac{\pi(a+1)}{n}\right)-\left(\psi\left(1-\dfrac{a+1}{n}\right)+\gamma\right)\right]\)

Q.E.D.

Ishan Singh - 2 years, 4 months ago

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

Ishan Singh - 2 years, 4 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 - 2 years, 4 months ago

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so LoveLy As always

Aman Rajput - 2 years, 5 months ago

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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 - 2 years, 5 months ago

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@Hasan Kassim

Kartik Sharma - 2 years, 5 months ago

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