Integral imprópria

por gsr_principiamathematica Qua 27 Dez 2023, 21:59

Prove que [latex]\int_{-\infty }^{\infty }\frac{1}{\sqrt{1+cosh(x)}}dx[/latex] é igual a [latex]\pi\cdot \sqrt{2}[/latex].

por **Giovana Martins** Qua 27 Dez 2023, 23:12

[latex]\mathrm{Identidade:cosh^2\left ( \frac{x}{2} \right )=\frac{1+cosh(x)}{2}}[/latex]

[latex]\mathrm{\int_{-\infty}^{+\infty } \frac{1}{\sqrt{1+cosh(x)}}dx=\int_{0}^{+\infty }\frac{2}{\sqrt{1+cosh(x)}}dx=\int_{0}^{+\infty }\frac{2}{\sqrt{2}cosh\left ( \frac{x}{2} \right )}dx}[/latex]

[latex]\mathrm{Seja\ \theta =\frac{x}{2}\ \therefore\ d\theta =\frac{1}{2}dx\ \therefore\ \int_{-\infty}^{+\infty } \frac{1}{\sqrt{1+cosh(x)}}dx=\int_{0}^{+\infty }\frac{2\sqrt{2}}{cosh(\theta )}d\theta }[/latex]

[latex]\mathrm{\int_{-\infty}^{+\infty } \frac{1}{\sqrt{1+cosh(x)}}dx=2\sqrt{2}\int_{0}^{+\infty }sech(\theta )d\theta =2\sqrt{2}arctan[sinh(\theta )]}[/latex]

[latex]\mathrm{\therefore\ \int_{-\infty}^{+\infty } \frac{1}{\sqrt{1+cosh(x)}}dx=2\sqrt{2}\left [ \lim_{t\to +\infty }arctan\left [sinh\left (\frac{t}{2} \right ) \right ]-\lim_{t\to 0 }arctan\left [sinh\left (\frac{t}{2} \right ) \right ] \right ]=\pi \sqrt{2}}[/latex]

A propósito, segue uma consideração que talvez possa gerar dúvidas acerca da quarta linha da resolução:

[latex]\\\mathrm{\ \ \ \ \ \ \ \int sech(\theta )d\theta =\int \frac{cosh(\theta )}{cosh^2(\theta )}d\theta =\int \frac{cosh(\theta )}{1+sinh^2(\theta )}d\theta }\\\\ \mathrm{u=sinh(\theta )\ \therefore\ d\theta =\frac{1}{cosh(\theta )}du\ \therefore\ \int sech(\theta )d\theta =\int \frac{1}{u^2+1}du}\\\\ \mathrm{\ \ \ \ \ \ \ \therefore\ \int sech(\theta )d\theta =arctan(u)=arctan(sinh(\theta ))+C}[/latex]

Integral imprópria

Integral imprópria

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