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Arithmetic Questions

Question Number 120654 by peter frank last updated on 01/Nov/20

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left[\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\right]^{\mathrm{n}} } \\$$

Answered by mindispower last updated on 01/Nov/20

$${x}={sh}\left({t}\right),{n}>\mathrm{1} \\$$ $$\Leftrightarrow\int_{\mathrm{0}} ^{\infty} \frac{{ch}\left({t}\right){dt}}{\left({ch}\left({t}\right)+{sh}\left({t}\right)\right)^{{n}} } \\$$ $$=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{t}} +{e}^{−{t}} }{\mathrm{2}{e}^{{nt}} }{dt} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{e}^{\left(\mathrm{1}−{n}\right){t}} }{\mathrm{1}−{n}}−\frac{{e}^{−\left(\mathrm{1}+{n}\right){t}} }{\mathrm{1}+{n}}\right]_{\mathrm{0}} ^{\infty} \: \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{n}−\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right]=\frac{{n}}{{n}^{\mathrm{2}} −\mathrm{1}} \\$$

Commented bypeter frank last updated on 01/Nov/20

$$\mathrm{thank}\:\mathrm{you} \\$$

Commented bymindispower last updated on 03/Nov/20

$${withe}\:{pleasur} \\$$

Answered by MJS_new last updated on 01/Nov/20

$$\int\frac{{dx}}{\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{{n}} }= \\$$ $$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dt}\right] \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{{n}+\mathrm{2}} }{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{t}^{{n}+\mathrm{2}} }+\frac{\mathrm{1}}{{t}^{{n}} }{dt}= \\$$ $$=−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right){t}^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right){t}^{{n}−\mathrm{1}} }=\:\left[\Rightarrow\right] \\$$ $$=−\frac{\left({n}+\mathrm{1}\right){t}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)}{\mathrm{2}\left({n}^{\mathrm{2}} −\mathrm{1}\right){t}^{{n}+\mathrm{1}} }= \\$$ $$=\frac{\left({x}+{n}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left(−{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{{n}} }{\mathrm{1}−{n}^{\mathrm{2}} }+{C} \\$$ $$\left[\Rightarrow\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{{n}} }=\right. \\$$ $$=\left[−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right){t}^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right){t}^{{n}−\mathrm{1}} }\right]_{\mathrm{1}} ^{\infty} = \\$$ $$=\frac{{n}}{\mathrm{1}−{n}^{\mathrm{2}} }\:\mathrm{with}\:{n}\in\mathbb{Z}\backslash\left\{\pm\mathrm{1}\right\} \\$$

Commented bypeter frank last updated on 01/Nov/20

$$\mathrm{thank}\:\mathrm{you} \\$$

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