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Question Number 195203 by Erico last updated on 27/Jul/23

$$\mathrm{Calculer}{\underset{0}{\int }}^{+\mathrm{\infty }}\frac{\mathrm{dt}}{{({\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}})}^2+{\mathrm{a}}^2}$$

Answered by Frix last updated on 27/Jul/23

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{{({\mathrm{e}}^t-{\mathrm{e}}^{-t})}^2+a^2}\stackrel{[u={\mathrm{e}}^{2t}]}{=}$$$$=\frac{1}{2}\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{du}{u^2+(a^2-2)u+1}=$$$$=\frac{1}{2a\sqrt{a^2-4}}{[\ln \mid \frac{2u+a^2-2+a\sqrt{a^2-4}}{2u+a^2-2-a\sqrt{a^2-4}}\mid ]}_1^{\mathrm{\infty }}=$$$$=\frac{1}{2a\sqrt{a^2-4}}\ln \mid \frac{a-\sqrt{a^2-4}}{a+\sqrt{a^2-4}}\mid$$

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