justify-that-0-dt-1-t-4-is-convergent-

Question Number 171071 by mathocean1 last updated on 07/Jun/22

j u s t i f y t h a t \int_{0}^{+ \infty} \frac{d t}{1 + t^{4}} i s c o n v e r g e n t .

Answered by aleks041103 last updated on 07/Jun/22

\int_{0}^{\infty} \frac{d t}{1 + t^{4}} = \int_{0}^{1} \frac{d t}{1 + t^{4}} + \int_{1}^{\infty} \frac{d t}{1 + t^{4}}

1 + t^{4} ⩾ 1 f o r 0 ⩽ t ⩽ 1

\Rightarrow \frac{1}{1 + t^{4}} ⩽ 1 \Rightarrow \int_{0}^{1} \frac{d t}{1 + t^{4}} < \int_{0}^{1} 1 d t = 1

\Rightarrow \int_{0}^{\infty} \frac{d t}{1 + t^{4}} < 1 + \int_{1}^{\infty} \frac{d t}{1 + t^{4}}

f o r t^{4} ⩾ t^{2} f o r t ⩾ 1

\Rightarrow \frac{1}{1 + t^{4}} ⩽ \frac{1}{1 + t^{2}}

\Rightarrow \int_{1}^{\infty} \frac{d t}{1 + t^{4}} ⩽ \int_{1}^{\infty} \frac{d t}{1 + t^{2}} = a r c t g {(t)}_{1}^{\infty} = \frac{π}{4}

\Rightarrow \int_{0}^{\infty} \frac{d t}{1 + t^{4}} < 1 + \frac{π}{4}

o b v .

\frac{1}{1 + t^{4}} > 0 \Rightarrow \int_{0}^{\infty} \frac{d t}{1 + t^{4}} > 0

\Rightarrow 0 < \int_{0}^{\infty} \frac{d t}{1 + t^{4}} < 1 + π / 4

\Rightarrow \int_{0}^{\infty} \frac{d t}{1 + t^{4}} c o n v e r g e s