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How can you show that $I=J=\phi^{-2}$

Where $\phi=\frac{\sqrt{5}+1}{2}$ $$I=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{4x^4+5x^2+1}\mathrm dx$$

$$J=\int_{-\infty}^{+\infty}\frac{\sin^2(4x\ln\phi)}{4x^4+5x^2+1}\mathrm dx$$

I don't think there are such formulas to represent $\sin^2(a)+\sin^2(b)=\sin^2(c)$

$$I=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{4x^4+5x^2+1}\mathrm dx=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{(4x^2+1)(x^2+1)}\mathrm dx$$

$$I=\int_{-\infty}^{+\infty}\frac{4\sin^2(x\ln\phi)+4\sin^2(3x\ln\phi)}{3(4x^2+1)}\mathrm dx-\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(x\ln\phi)}{3(x^2+1)}\mathrm dx$$

In general we got evalaute, $$K=\int_{-\infty}^{+\infty}\frac{\sin^2(ax\ln\phi)}{bx^2+c}\mathrm dx$$

$\sin^2(a)=\frac{1-\cos(2a)}{2}$

$$K=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{1-\cos(2ax\ln\phi)}{bx^2+c}\mathrm dx=\frac{\pi}{2b}\sqrt{\frac{b}{c}}-\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\cos(2ax\ln\phi)}{bx^2+c}\mathrm dx$$

Sibawayh
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2 Answers2

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You can use this result $$ \int_{-\infty}^\infty \frac{\cos x}{x^2+a^2}dx=\frac{e^{-a}}{a}\pi$$ to evaluate $$ \int_{-\infty}^\infty \frac{\cos (ax)}{b^2x^2+c^2}dx=\frac1a\int_{-\infty}^\infty \frac{\cos (x)}{\frac{b^2}{a^2}x^2+c^2}dx=\frac{a}{b^2}\int_{-\infty}^\infty \frac{\cos (x)}{x^2+\frac{a^2c^2}{b^2}}dx=\frac{a}{b^2}\frac{e^{-\frac{ac}{b}}}{\frac{ac}{b}}\pi=\frac{\pi}{bc}e^{-\frac{ac}{b}} $$ For example \begin{eqnarray} &&\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)}{4x^4+5x^2+1}\mathrm dx\\ &=&-\frac13\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)}{x^2+1}\mathrm dx+\frac43\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)}{4x^2+1}\mathrm dx. \end{eqnarray} But $$\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)}{x^2+1}\mathrm dx=\frac12\int_{-\infty}^{+\infty}\frac{1}{x^2+1}\mathrm dx-\frac12\int_{-\infty}^{+\infty}\frac{\cos(2x\ln\phi)}{x^2+1}\mathrm dx=\frac12\pi-\frac12\cdot\pi e^{-2\ln\phi}=\frac12\pi-\frac12\pi\frac{1}{\phi^2}$$ and you can use the same way to handle the rest.

xpaul
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The key integral given at the start of @xpaul's answer can be obtained using algebra $$\int\frac{\cos (x)}{x^2+a^2}\,dx=\int\frac{\cos (x)}{(x+ia)(x-ia)}\,dx$$ $$I=\int \frac{i\,\cos(x)}{2 a (x+i a)}\,dx-\int\frac{i\,\cos(x)}{2 a (x-i a)}\, dx$$ $$I=\frac{i \cosh (a) \text{Ci}(i a+x)-\sinh (a) \text{Si}(i a+x)}{2 a}+$$ $$\frac{\sinh (a) \text{Si}(i a-x)-i \cosh (a) \text{Ci}(x-i a)}{2 a}$$

Assuming $a>0$ $$\int_{-\infty}^\infty \frac{\cos x}{x^2+a^2}dx=\frac{\pi}{a} e^{-a}$$

Claude Leibovici
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