A simple integral, II

Here is problem #675 from The Pentagon magazine published by Kappa Mu Epsilon. The problem and the solution below are by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Let \(k\) be a real number with \(k \gt -1\). Calculate the double integral

\(\displaystyle\int_{0}^{1}\int_{0}^{1}\frac{x^{k+1}y^{k}}{x+y}dxdy.\)

There is a way to find this integral actually without computing integrals at all. Well, almost.

Let \(k\) be a real number with \(k \gt -1\). Calculate the double integral

\(\displaystyle\int_{0}^{1}\int_{0}^{1}\frac{x^{k+1}y^{k}}{x+y}dxdy.\)

Solution

Let \(I\) denote the value of the integral. By symmetry, we have

\(\displaystyle I=\int_{0}^{1}\int_{0}^{1}\frac{x^{k+1}y^{k}}{x+y}dxdy=\int_{0}^{1}\int_{0}^{1}\frac{x^{k}y^{k+1}}{x+y}dxdy.\)

Therefore,

\( \displaystyle \begin{array}{4,1} I &= \frac{1}{2}(I + I) \\ &= \frac{1}{2}(\int_{0}^{1}\int_{0}^{1}\frac{x^{k+1}y^{k}}{x+y}dxdy+\int_{0}^{1}\int_{0}^{1}\frac{x^{k}y^{k+1}}{x+y}dxdy) \\ &= \frac{1}{2}\int_{0}^{1}\int_{0}^{1}{x^{k}y^{k}}dxdy \\ &= \frac{1}{2}\frac{1}{(k+1)^{2}}. \end{array} \)

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