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Question Number 171071 by mathocean1 last updated on 07/Jun/22
justify that ∫_0 ^(+∞) (dt/(1+t^4 )) is convergent.
$${justify}\:{that}\:\int_{\mathrm{0}} ^{+\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{is}\:{convergent}. \\ $$
Answered by aleks041103 last updated on 07/Jun/22
∫_0 ^∞ (dt/(1+t^4 ))=∫_0 ^1 (dt/(1+t^4 )) + ∫_1 ^∞ (dt/(1+t^4 )) 1+t^4 ≥1 for 0≤t≤1 ⇒(1/(1+t^4 ))≤1⇒∫_0 ^1 (dt/(1+t^4 ))<∫_0 ^1 1dt=1 ⇒∫_0 ^∞ (dt/(1+t^4 ))<1+∫_1 ^∞ (dt/(1+t^4 )) for t^4 ≥t^2 for t≥1 ⇒(1/(1+t^4 ))≤(1/(1+t^2 )) ⇒∫_1 ^∞ (dt/(1+t^4 ))≤∫_1 ^∞ (dt/(1+t^2 ))=arctg(t)_1 ^∞ =(π/4) ⇒∫_0 ^∞ (dt/(1+t^4 ))<1+(π/4) obv. (1/(1+t^4 ))>0⇒∫_0 ^∞ (dt/(1+t^4 ))>0 ⇒0<∫_0 ^∞ (dt/(1+t^4 ))<1+π/4 ⇒∫_0 ^∞ (dt/(1+t^4 )) converges
$$\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:+\:\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\mathrm{1}+{t}^{\mathrm{4}} \geqslant\mathrm{1}\:{for}\:\mathrm{0}\leqslant{t}\leqslant\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }\leqslant\mathrm{1}\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }<\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}{dt}=\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }<\mathrm{1}+\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$${for}\:{t}^{\mathrm{4}} \geqslant{t}^{\mathrm{2}} \:{for}\:{t}\geqslant\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }\leqslant\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\leqslant\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }={arctg}\left({t}\right)_{\mathrm{1}} ^{\infty} =\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }<\mathrm{1}+\frac{\pi}{\mathrm{4}} \\ $$$${obv}. \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }>\mathrm{0}\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }>\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}<\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }<\mathrm{1}+\pi/\mathrm{4} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{converges} \\ $$

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