 Algotssleeddynf

2022-01-04

I have to compute this integral
${\int }_{0}^{\mathrm{\infty }}\frac{dt}{1+{t}^{4}}$ Alex Sheppard

Note that the substitution $t=\frac{1}{u}$ changes the integral to
${\int }_{0}^{\mathrm{\infty }}\frac{{u}^{2}}{1+{u}^{4}}du$
Split the integral into into two parts, 0 to 1, and 1 to infinity. On the second part, let $t=\frac{1}{u}$. We get ${\int }_{0}^{1}\frac{{u}^{2}}{1+{u}^{4}}du$. Now u has done its duty, and is discarded for the more popular t. Our original integral is equal to
${\int }_{0}^{1}\frac{1+{t}^{2}}{1+{t}^{4}}dt$
There is now a minor miracle.
$\frac{1}{1-\sqrt{2}t+{t}^{2}}+\frac{1}{1+\sqrt{2}t+{t}^{2}}=\frac{2\left(1+{t}^{2}\right)}{1+{t}^{4}}$
Complete the square(s) as usual. karton

Let the considered integral be I i.e
$I={\int }_{0}^{\mathrm{\infty }}\frac{1}{1+{t}^{4}}dt$
Under the transformation $t↦1/t$, the integral is:
$I={\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}}{1+{t}^{4}}dt⇒2I={\int }_{0}^{\mathrm{\infty }}\frac{1+{t}^{2}}{1+{t}^{4}}dt={\int }_{0}^{\mathrm{\infty }}\frac{1+\frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}}dt$
$2I={\int }_{0}^{\mathrm{\infty }}\frac{1+\frac{1}{{t}^{2}}}{\left(t-\frac{1}{t}{\right)}^{2}+2}dt$
Next, use the substitution $1-t/t=u⇒\left(1+1/{t}^{2}\right)dt=du$ to get:
$2I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{du}{{u}^{2}+2}⇒I={\int }_{0}^{\mathrm{\infty }}\frac{du}{{u}^{2}+2}=\frac{\pi }{2\sqrt{2}}$ user_27qwe

If you don't want to do a partial fraction decomposition, put $I:={\int }_{0}^{+\mathrm{\infty }}\frac{dt}{1+{t}^{4}}$. By the substitution $x=\frac{1}{t}$ on $\left[0,+\mathrm{\infty }\right)$ and $\left(-\mathrm{\infty },0\right]$, we get ${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{x}^{2}}{1+{x}^{4}}dx$. Now we have
$4I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{t}^{2}-\sqrt{2}t+1}{\left({t}^{2}-\sqrt{2}t+1\right)\left({t}^{2}+\sqrt{2}t+1\right)}dt\phantom{\rule{0ex}{0ex}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{t}^{2}+\sqrt{2}t+1}dt\phantom{\rule{0ex}{0ex}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{\left(t+\frac{\sqrt{2}}{2}{\right)}^{2}-\frac{1}{2}+1}dt\phantom{\rule{0ex}{0ex}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{du}{{u}^{2}+\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{du}{\left(\sqrt{2}u{\right)}^{2}+1}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{2}}\mathrm{arctan}\left(\sqrt{2}u\right){|}_{u=-\mathrm{\infty }}^{u=+\mathrm{\infty }}\phantom{\rule{0ex}{0ex}}=\frac{2\pi }{\sqrt{2}}$
and finally $I=\frac{2\pi }{\sqrt{2}}$

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