Following is the original integral: $$F(m,n)=\displaystyle \int_{0}^{1}{\dfrac{\arctan(x^{m})\arctan(x^{n})}{x}\ \mathrm{d}x}$$ Then, the bivariate function $F(m,n)$ is given the second partial derivative, first m and then n. Like: $$\dfrac{\partial^2F(m,n)}{\partial m\partial n}=\dfrac{1}{4}\int_{0}^{\infty}{x^2\ \mathrm{sech}(mx)\ \mathrm{sech}(nx)\ \mathrm{d}x}$$ I guess Merlin transformation or Ramanujan's master theorem is needed. However, I can't find the right way.
2026-03-25 17:19:50.1774459190
An integral of $\displaystyle \int_{0}^{\infty}{x^2\ \mathrm{sech}(mx)\ \mathrm{sech}(nx)\ \mathrm{d}x}$, m,n $∈{\mathbb{R}_{+}}$
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