I have to compute this integral $$\int_{0}^{\infty} \frac{dt}{1+t^4}$$ to solve a problem in a homework. I have tried in many ways, but I'm stuck. A search in the web reveals me that it can be do it by methods of complex analysis. But I have not taken this course yet. Thanks for any help.
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- 10$\begingroup$ Complex Analysis is the best way to go about. Else write $1+t^4 = (1 + \sqrt{2} t + t^2 ) \times (1 - \sqrt{2} t + t^2 )$ and do partial fractions. $\endgroup$user17762– user177622011-06-05 19:18:02 +00:00Commented Jun 5, 2011 at 19:18
- 1$\begingroup$ More solutions can be found in the following two questions: one and two. $\endgroup$user642796– user6427962015-01-09 04:53:05 +00:00Commented Jan 9, 2015 at 4:53
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8 Answers
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2 Note that the substitution $t=1/u$ changes the integral to $$\int_0^\infty \frac{u^2}{1+u^4}du.$$ Doesn't sound very helpful, but there was extensive discussion of that here.
Added: But we can do better. Split the integral into into two parts, $0$ to $1$, and $1$ to infinity. On the second part, let $t=1/u$. 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. $$\frac{1}{1-\sqrt{2}t+t^2}+ \frac{1}{1+\sqrt{2}t+t^2}=\frac{2(1+t^2)}{1+t^4}.$$
Complete the square(s) as usual.
- 20$\begingroup$ A nicer miracle: $ \displaystyle \int^1_0 \frac{1+t^2}{1+t^4} dt = \int^1_0 \frac{d(t-1/t) }{(t-1/t)^2+2} = \frac{1}{\sqrt{2}} \arctan \left( \frac{t-1/t}{\sqrt{2}} \right) \Bigg|^1_0 = \frac{\pi}{2\sqrt{2} }$ $\endgroup$Ragib Zaman– Ragib Zaman2011-11-15 06:45:14 +00:00Commented Nov 15, 2011 at 6:45
- $\begingroup$ Indeed it is nicer, symmetrizes in one swoop. $\endgroup$André Nicolas– André Nicolas2011-11-15 06:54:26 +00:00Commented Nov 15, 2011 at 6:54
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3 If you don't want to do a partial fraction decomposition, put $I:=\int_0^{+\infty}\frac{dt}{1+t^4}$. By the substitution $x=\frac 1t$ on $\left[0,+\infty\right)$ and $\left(-\infty,0\right]$, we get $2I = \int_{-\infty}^{+\infty}\frac{x^2}{1+x^4}\,dx$. Now we have \begin{align*} 4I &= \int_{-\infty}^{+\infty}\frac{t^2-\sqrt 2t+1}{(t^2-\sqrt 2t+1)(t^2+\sqrt 2t+1)}\,dt \\\ &=\int_{-\infty}^{+\infty}\frac{1}{t^2+\sqrt 2t+1}\,dt\\\ &=\int_{-\infty}^{+\infty}\frac{1}{(t+\frac{\sqrt 2}2)^2-\frac 12+1}\,dt\\\ &=\int_{-\infty}^{+\infty}\frac{du}{u^2+\frac 12}\\\ &=2\int_{-\infty}^{+\infty}\frac{du}{(\sqrt 2 u)^2+1}\\\ &=\frac 2{\sqrt 2}\arctan (\sqrt 2u)\mid_{u=-\infty}^{u=+\infty}\\\ &=\frac {2\pi}{\sqrt 2} \end{align*} and finally $I=\dfrac{\pi}{2\sqrt 2}$.
- $\begingroup$ Why $4I$ is that you claim, in your first equality? $\endgroup$leo– leo2011-06-05 20:40:22 +00:00Commented Jun 5, 2011 at 20:40
- 3$\begingroup$ We have $2I =\int_{-\infty}^{+\infty}\frac{t^2}{1+t^4}dt =\int_{-\infty}^{+\infty}\frac{dt}{1+t^4}$, and I added these two terms. $\endgroup$Davide Giraudo– Davide Giraudo2011-06-05 20:45:09 +00:00Commented Jun 5, 2011 at 20:45
- $\begingroup$ ok. It's fine. Thank you. $\endgroup$leo– leo2011-06-05 21:32:40 +00:00Commented Jun 5, 2011 at 21:32
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{x \equiv {1 \over 1 + t^{4}} \quad\iff\quad t = \pars{1 - x \over x}^{1/4}}$: \begin{align} & \color{#00f}{\large\int_{0}^{\infty}{\dd t \over 1 + t^{4}}} = \int_{1}^{0}x\,{1 \over 4}\,\pars{1 - x \over x}^{-3/4} \pars{-\,{1 \over x^{2}}}\,\dd x \\[5mm] = & \ {1 \over 4}\int_{0}^{1}x^{-1/4}\,\, \pars{1 - x}^{-3/4}\,\,\dd x = {1 \over 4}\,{\rm B}\pars{{3 \over 4},{1 \over 4}} \\[5mm] = & \ {1 \over 4}\,{\Gamma\pars{3/4}\Gamma\pars{1/4} \over \Gamma\pars{3/4 + 1/4}} ={1 \over 4}\,{\pi \over \sin\pars{\pi\bracks{1/4}}} \\[5mm] = & \ \begin{array}{|c|}\hline \rule{0pt}{1cm}\color{#00f}{\large{\root{2} \over 4}\,\pi} \\ \hline \end{array} \approx 1.1107 \end{align}
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Let the considered integral be I i.e $$I=\int_0^{\infty} \frac{1}{1+t^4}\,dt$$ Under the transformation $t\mapsto 1/t$, the integral is: $$I=\int_0^{\infty} \frac{t^2}{1+t^4}\,dt \Rightarrow 2I=\int_0^{\infty}\frac{1+t^2}{1+t^4}\,dt=\int_0^{\infty} \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\,dt$$ $$2I=\int_0^{\infty} \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2}\,dt$$ Next, use the substitution $t-1/t=u \Rightarrow (1+1/t^2)\,dt=du$ to get: $$2I=\int_{-\infty}^{\infty} \frac{du}{u^2+2}\Rightarrow I=\int_0^{\infty} \frac{du}{u^2+2}=\boxed{\dfrac{\pi}{2\sqrt{2}}}$$ $\blacksquare$
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You can also do it by partial fraction decomposition. We have $$ \frac{2\sqrt{2}}{1+t^4} = \frac{t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{t - \sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$ We have $$\int_0^\infty \frac{dt}{t^2 \pm \sqrt{2}t + 1} = \int_0^\infty \frac{dt}{(t \pm \sqrt{2}/2)^2 + 1/2} = \int_0^\infty \frac{2dt}{(\sqrt{2} t \pm 1)^2 + 1} = \sqrt{2} \arctan (\sqrt{2}t \pm 1) \big|_0^\infty.$$ Continuing, we get $$\frac{\sqrt{2}\pi}{2} - \sqrt{2} \arctan (\pm 1) = \frac{\sqrt{2}\pi}{2} \mp \frac{\sqrt{2}}{4} = \frac{\sqrt{2}\pi (2 \mp 1)}{4}.$$ Next, we have $$ \int_0^\infty \frac{(2t \pm \sqrt{2}) dt}{t^2 \pm \sqrt{2}t + 1} = \log (t^2 \pm \sqrt{2}t + 1) \big|_0^\infty. $$ The integral doesn't converge, but we can consider instead $$ \int_0^\infty \frac{(2t + \sqrt{2}) dt}{t^2 + \sqrt{2}t + 1} - \frac{(2t - \sqrt{2}) dt}{t^2 - \sqrt{2}t + 1} = \log \frac{t^2 + \sqrt{2}t + 1}{t^2 - \sqrt{2}t + 1} \big|_0^\infty = 0. $$ Therefore we rewrite our initial expansion $$ \frac{4\sqrt{2}}{1+t^4} = \frac{2t + 2\sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{2t - 2\sqrt{2}}{t^2 - \sqrt{2}t + 1} = \frac{2t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{2t - \sqrt{2}}{t^2 - \sqrt{2}t + 1} + \frac{\sqrt{2}}{t^2 + \sqrt{2}t + 1} + \frac{\sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$ Integrating, we get $$ \int_0^\infty \frac{4\sqrt{2}}{1+t^4} = \sqrt{2} \frac{\sqrt{2} \pi (2-1)}{4} + \sqrt{2} \frac{\sqrt{2} \pi (2+1)}{4} = \frac{\pi}{2} + \frac{3\pi}{2} = 2\pi. $$ Therefore the integral we want is $$ \int_0^\infty \frac{1}{1 + t^4} = \frac{2\pi}{4\sqrt{2}} = \frac{\sqrt{2}\pi}{4}. $$
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0 $1 + t^4 = (1 + 2t^2 + t^4) - 2t^2 = (1 + t^2)^2 - (\sqrt{2} t)^2$
This is a difference of squares and so can be factored. Once you've factored it, use partial fractions.
If you don't know that trick, you can do this:
$1+t^4=0$ iff $t^4 = -1$, so $t^2 = \pm i$. If $t^2 = i$, then $t = \pm \frac{1+i}{\sqrt{2}}$, and if $t^2 = -i$, then $t = \pm\frac{1-i}{\sqrt{2}}$. The way you get this is that when you multiply complex numbers, you add angles and multiply lengths; hence when you take the square root of a complex number, you take the square root of the length and cut the angle in half.
So now you've got $1+t^4 = \big( (t - (1+i)/\sqrt{2})(t - (1-i)/\sqrt{2})\big)\big((t - (-1+i)/\sqrt{2})(t - (-1-i)/\sqrt{2})\big)$. If you multiply the first two factors, you get $t^2 - t\sqrt{2} + 1$. If you multiply the last two factors, you get $t^2 + t\sqrt{2} + 1$.
Once you've factored the denominator, use partial fractions. You've got an irreducible quadratic factor in the denominator, so you'll get an arctangent.
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1 $$ \begin{aligned} \int_0^{\infty} \frac{d t}{1+t^4}&=\int_0^{\infty} \frac{\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t \\=& \frac{1}{2} \int_0^{\infty} \frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}} d t \\ =& \frac{1}{2}\left[\int_0^{\infty} \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+2}-\int_0^{\infty} \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2-2}\right]_0^{\infty} \\ =& \frac{1}{2 \sqrt{2}}\left[\tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)\right]_0^{\infty}-\frac{1}{2 \sqrt{2}}\left[\ln \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}} \right|\right]_0^{\infty}\\ =& \frac{\pi}{2 \sqrt{2}} \end{aligned} $$
- 2$\begingroup$ The starting integral should have the power of $t$ as $4$ and not $2$. $\endgroup$Poonguzhali Annadurai– Poonguzhali Annadurai2025-05-10 11:17:46 +00:00Commented May 10 at 11:17
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\begin{align} \int_0^{\infty}\frac{dt}{1+t^4} &=\int_0^{\infty}\int_0^{\infty}e^{-s(1+t^4)}\,ds\,dt & \\ &=\int_0^{\infty}e^{-s}\int_0^{\infty}e^{-st^4}\,dt\,ds & \left[t=\left(\frac{x}{s}\right)^{1/4}\right] \\ &=\frac{1}{4}\int_0^{\infty}s^{-1/4}e^{-s}\,ds\int_0^{\infty}x^{-3/4}e^{-x}\,dx & \left[\Gamma(z)=\int_0^{\infty}t^{z-1}e^{-t}\,dt\right]\\ &=\frac{1}{4}\Gamma\!\left(\frac{3}{4}\right)\Gamma\!\left(\frac{1}{4}\right) & \left[\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}\right] \\ &=\frac{\pi}{4\sin\left(\frac{\pi}{4}\right)} & \\ &=\frac{\pi}{2\sqrt{2}} \end{align}