Solution of Problem 1

Please solve Problem 1 .

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First, we discuss $A(k)=B(k)$. One sets $y=\f{1}{x}$, $dy=-\f{dx}{x^2}=-y^2dx$ and $dx=-\f{dy}{y^2}$, then,
\begin{eqnarray} A(k)\eq\int_0^\infty\f{x^2+kx+1}{x^4+1}\tan^{-1}\left(\f{1}{x}\right)dx\nn \eq\int_\infty^0\f{\f{1}{y^2}+\f{k}{y}+1}{\f{1}{y^4}+1}\tan^{-1}y\left(-\f{dy}{y^2}\right)\nn \eq\int_0^\infty\f{1+ky+y^2}{1+y^4}\tan^{-1}ydy=B(k). \end{eqnarray}

is derived. Moreover, using $\tan^{-1}a+\tan^{-1}\f{1}{a}=\f{\pi}{2}$, we summarize $A(k)+B(k)$ as

\begin{eqnarray} A(k)+B(k)\eq \int_0^\infty\f{x^2+kx+1}{x^4+1}\left\{\tan^{-1}\left(\f{1}{x}\right)+\tan^{-1}(x)\right\}dx\nn \eq\f{\pi}{2} \int_0^\infty\f{x^2+kx+1}{x^4+1}dx. \end{eqnarray}

Thus,
\begin{eqnarray} A(k)\eq\f{A(k)+B(k)}{2}= \f{\pi}{4} \int_0^\infty\f{x^2+kx+1}{x^4+1}dx, \end{eqnarray}
is obtained. In order to integrate, we can factorize $x^4+1$ as $x^4+2x^2+1-2x^2$ $=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$, then,
\begin{eqnarray} \f{x^2+kx+1}{x^4+1}\eq \f{a}{x^2+\sqrt{2}x+1}+\f{b}{x^2-\sqrt{2}x+1}\nn \eq\f{a(x^2-\sqrt{2}x+1)+b(x^2+\sqrt{2}x+1)} {(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}\nn \eq\f{(a+b)(x^2+1)+\sqrt{2}(b-a)x}{x^4+1}. \end{eqnarray}
From $a+b=1,\sqrt{2}(b-a)=k$, we get
\begin{eqnarray} a\eq\f{1}{2}-\f{k}{2\sqrt{2}},\\ b\eq\f{1}{2}+\f{k}{2\sqrt{2}}. \end{eqnarray}

Therefore,
\begin{eqnarray} \int_0^\infty\f{x^2+kx+1}{x^4+1}dx \eq\left(\f{1}{2}-\f{k}{2\sqrt{2}}\right)\int_0^\infty\f{dx}{x^2+\sqrt{2}x+1}\nn &&+\left(\f{1}{2}+\f{k}{2\sqrt{2}}\right)\int_0^\infty\f{dx}{x^2-\sqrt{2}x+1}, \end{eqnarray}
is calculated. We use $x+\f{1}{\sqrt{2}}=\f{1}{\sqrt{2}}\tan\theta$ and $dx=\f{d\theta}{\sqrt{2}\cos^2\theta}$ so as to integrate, the first term is calculated as
\begin{eqnarray} \int_0^\infty \f{dx}{x^2+\sqrt{2}x+1}\eq \int_0^\infty \f{dx}{\left(x+\f{1}{\sqrt{2}}\right)^2+\f{1}{2}} \nn \eq\int_{\f{\pi}{4}}^{\f{\pi}{2}}\f{d\theta}{\sqrt{2}\cos^2\theta}\f{2}{\tan^2\theta+1}\nn \eq\f{2}{\sqrt{2}}\int_{\f{\pi}{4}}^{\f{\pi}{2}}d\theta=\f{2}{\sqrt{2}}\left(\f{\pi}{2}-\f{\pi}{4}\right)=\f{\pi}{2\sqrt{2}}. \end{eqnarray}
Next, we use $x-\f{1}{\sqrt{2}}=\f{1}{\sqrt{2}}\tan\theta$ and $dx=\f{d\theta}{\sqrt{2}\cos^2\theta}$ in order to integrate, the second term is evaluated as
\begin{eqnarray} \int_0^\infty \f{dx}{x^2-\sqrt{2}x+1}\eq \int_0^\infty \f{dx}{\left(x-\f{1}{\sqrt{2}}\right)^2+\f{1}{2}} \nn \eq\int_{-\f{\pi}{4}}^{\f{\pi}{2}}\f{d\theta}{\sqrt{2}\cos^2\theta}\f{2}{\tan^2\theta+1}\nn \eq\f{2}{\sqrt{2}}\int_{-\f{\pi}{4}}^{\f{\pi}{2}}d\theta=\f{2}{\sqrt{2}}\left(\f{\pi}{2}-\left(-\f{\pi}{4}\right)\right)=\f{3\pi}{2\sqrt{2}}.\qquad \end{eqnarray}
Therefore, the final result is
\begin{eqnarray} A(k)\eq\f{\pi}{4}\int_0^\infty\f{x^2+kx+1}{x^4+1}dx\nn \eq\f{\pi}{4}\left(\f{1}{2}-\f{k}{2\sqrt{2}}\right)\int_0^\infty\f{dx}{x^2+\sqrt{2}x+1}\nn &&+\f{\pi}{4}\left(\f{1}{2}+\f{k}{2\sqrt{2}}\right)\int_0^\infty\f{dx}{x^2-\sqrt{2}x+1}\nn \eq\f{\pi}{4}\left(\f{1}{2}-\f{k}{2\sqrt{2}}\right)\f{\pi}{2\sqrt{2}} + \f{\pi}{4}\left(\f{1}{2}+\f{k}{2\sqrt{2}}\right)\f{3\pi}{2\sqrt{2}}\nn \eq\f{\pi}{4}\f{\pi}{2\sqrt{2}}\left(\f{1}{2}+\f{3}{2}\right)+\f{\pi}{4}\f{\pi}{2\sqrt{2}}\left(-\f{1}{2\sqrt{2}}+\f{3}{2\sqrt{2}}\right)k\nn \eq\f{\pi^2\sqrt{2}}{8}+\f{\pi^2}{16}k, \end{eqnarray}
where $B(k)=\f{\pi^2\sqrt{2}}{8}+\f{\pi^2}{16}k$ is also derived.

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