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Question Number 125708 by Dwaipayan Shikari last updated on 13/Dec/20

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{tanhx}}{\:\sqrt[{\mathrm{5}}]{{tanhx}}}{dx} \\$$

Answered by mindispower last updated on 13/Dec/20

$${e}^{{x}} ={t} \\$$ $$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}−\frac{{t}−\frac{\mathrm{1}}{{t}}}{{t}+\frac{\mathrm{1}}{{t}}}}{\:\sqrt[{\mathrm{5}}]{\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}}}.\frac{\mathrm{1}}{{t}}{dt} \\$$ $$=\int_{\mathrm{1}} ^{\infty} \frac{\frac{\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt[{\mathrm{5}}]{\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}}}.\frac{{dt}}{{t}}=\mathrm{2}\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} } \\$$ $$=\int_{\mathrm{1}} ^{\infty} \frac{{d}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} }\: \\$$ $$=\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{{t}\left({t}+\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} \left({t}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} } \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{d}\left(\frac{\mathrm{1}}{{t}}\right)}{\frac{\mathrm{1}}{{t}}\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} .\frac{\mathrm{1}}{{t}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{t}\right)^{−\frac{\mathrm{4}}{\mathrm{5}}} \left(\mathrm{1}−{t}\right)^{−\frac{\mathrm{1}}{\mathrm{5}}} {dt} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{1}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{9}}{\mathrm{5}}−\mathrm{1}−\mathrm{1}} \left(\mathrm{1}−\left(−\mathrm{1}\right){t}\right)^{−\frac{\mathrm{4}}{\mathrm{5}}} {dt} \\$$ $$=\beta\left(\mathrm{1},\frac{\mathrm{4}}{\mathrm{5}}\right)\:\:\:\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{5}},\mathrm{1};\frac{\mathrm{9}}{\mathrm{5}};−\mathrm{1}\right) \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{th}\left({x}\right)}{\:\sqrt[{\mathrm{5}}]{{th}\left({x}\right)}}{dx}=\beta\left(\mathrm{1},\frac{\mathrm{4}}{\mathrm{5}}\right)._{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{5}},\mathrm{1};\frac{\mathrm{9}}{\mathrm{5}};\mathrm{1}\right)\backsimeq\mathrm{0}.\mathrm{90323} \\$$

Commented byDwaipayan Shikari last updated on 13/Dec/20

$${Great}\:{sir}!\:{great}\:{approach}\:. \\$$ $${My}\:{thinking} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{tanhx}}{\:\sqrt[{\mathrm{5}}]{{tanhx}}}{dx}=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{tanh}^{\mathrm{2}} {x}}{\left(\mathrm{1}+{tanhx}\right)\sqrt[{\mathrm{5}}]{{tanhx}}}{dx}\:\:\:\:{tanhx}={t} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left(\mathrm{1}+{t}\right)\sqrt[{\mathrm{5}}]{{t}}}=\mathrm{5}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{\mathrm{3}} }{\mathrm{1}+{u}^{\mathrm{5}} }{dt}\:....\:\:\:\:\:\:{t}={u}^{\mathrm{5}} \: \\$$ $${We}\:{can}\:{decompose}\:{it}\: \\$$

Commented bymindispower last updated on 13/Dec/20

$${yes}\:{nice} \\$$ $${nice} \\$$