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How could one solve $$\int_0^\infty \frac{1}{1-t^4} \, dt\,?$$ I have to apply special functions, so I thought that I have to use the change variable $$u=t^4,$$ but $$du=4t^3\,dt$$ and when $$t\rightarrow0\qquad u\rightarrow0\ $$ I get $$t\rightarrow\infty\qquad u\rightarrow\infty, $$

whereas beta function is $\int_{0}^{1} t^{x-1}(1-t)^{y-1}dt$ so I cannot use that change. Help?

Brightsun
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Cheter
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    Factorise and use partial fractions would be the standard method – Mark Bennet Jan 17 '16 at 16:16
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    $\dfrac{1}{1-t^4}$ has a pole of order $1$ at $1$, so the integral doesn't exist in a strict sense. You need to consider it as a principal value integral. – Daniel Fischer Jan 17 '16 at 16:17
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    Hint: consider $u = \dfrac{1}{t}$ (but don't forget the problem at $1$). – Daniel Fischer Jan 17 '16 at 16:20
  • Then $t\rightarrow 0; \ u\rightarrow \infty$, I need 0, then 1 :S, thanks for try help – Cheter Jan 17 '16 at 16:23
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    Aside from the pole at $1$, you can do the integral using partial fractions. What special function were you thinking to use? You get $\int 1/(1-x^4) dx = 1/4 (-\log(1-x)+\log(x+1)+2 \tan^{-1}(x))+c$ Alpha gives some other forms as well. – Ross Millikan Jan 17 '16 at 16:25
  • I think that I could use $beta$ function, which is defined by $\int_{0}^{1} t^{2x-1}(1-t)^{2y-1}dt$, When I use that $u=t^4$ I get $\int_{0}^{\infty} u^{3}(1-u)^{-1}du$ – Cheter Jan 17 '16 at 16:28
  • duplicate: http://math.stackexchange.com/questions/43457/how-can-i-compute-the-integral-int-0-infty-fracdt1t4 – 永劫回帰 Jan 17 '16 at 16:54
  • from above link (with gamma and beta): http://math.stackexchange.com/a/651818/110891 – 永劫回帰 Jan 17 '16 at 16:58
  • @変幻出没 Not a duplicate (but a useful link anyway): it's $\frac{1}{1\color{red}{+} t^4}$ in that question and $\frac{1}{1\color{red}{-} t^4}$ here. – Winther Jan 17 '16 at 17:04
  • Thanks too, @変幻出没 that change and brightsun give me a big idea to solve the problem!, thanks each one – Cheter Jan 17 '16 at 17:13
  • In general, $~\displaystyle\int_0^\infty\frac{x^{k-1}}{1-x^n}~dx ~=~ \frac\pi n~\cot\bigg(k~\frac\pi n\bigg).$ – Lucian Jan 17 '16 at 19:13

3 Answers3

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(Assuming principal value)

Let $t=u^4$ as you suggested, then we get $$ PV\int_{0}^{+\infty}\frac{du}{4u^{3/4}(1-u)}= \lim_{\varepsilon\to0}\left[\int_{0}^{1-\varepsilon}\frac{du}{4u^{3/4}(1-u)}+\int_{1+\varepsilon}^{+\infty}\frac{du}{4u^{3/4}(1-u)}\right] $$ and now an idea could be letting $u=1/(1-y)$ in the second integral to get $$ -\frac{1}{4}\int_{\varepsilon/(1+\varepsilon)}^{1}y^{-1}(1-y)^{-1/4}dy, $$ so $$ PV\int_{0}^{+\infty}\frac{du}{4u^{3/4}(1-u)}=\frac{1}{4} \lim_{\varepsilon\to0}\left[\int_{0}^{1-\varepsilon}u^{-3/4}(1-u)^{-1}du-\int_{\varepsilon/(1+\varepsilon)}^{1}y^{-1}(1-y)^{-1/4}dy\right]. $$ Now $$ \beta(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, $$ and since the original integral is not well defined, we get $+\infty-\infty$. Letting also $u=1-y$ in the first term, \begin{align} PV\int_{0}^{+\infty}\frac{dt}{1-t^4}&=\frac{1}{4} \lim_{\varepsilon\to0}\left[\int_{\varepsilon}^{1}(1-y)^{-3/4}y^{-1}dy-\int_{\varepsilon/(1+\varepsilon)}^{1}y^{-1}(1-y)^{-1/4}dy\right]\\ &=\frac{1}{4} \int_{0}^{1}\left[(1-y)^{-3/4}-(1-y)^{-1/4}\right]y^{-1}dy\\ &=\frac{1}{4} \int_{0}^{1}\frac{u^{-3/4}-u^{-1/4}}{1-u}du=\frac{1}{2}\int_{0}^{1}\frac{1-\sqrt u}{u^{1/4}(1-u)}\frac{du}{2\sqrt u}\\ &=\frac{1}{2}\int_{0}^{1}s^{-1/2}(1+s)^{-1}ds\\ &=\left[\tan^{-1}(\sqrt s)\right]_0^1=\pi/4, \end{align} where in the second step $$ \left| \int_{\varepsilon/(\varepsilon+1)}^\varepsilon y^{-1}(1-y)^{-1/4}dy\right|\le \frac{\varepsilon +1}{\varepsilon}\int_{\varepsilon/(\varepsilon+1)}^\varepsilon dy=\frac{\varepsilon+1}{\varepsilon}\frac{\varepsilon^2}{\varepsilon+1}\to0 $$ has been used.

If you admit complex methods in general, let $$ PV\int_0^{+\infty}\frac{dt}{1-t^4}=\frac{1}{2}PV\int_{-\infty}^{+\infty}\frac{dt}{1-t^4}. $$ Now, using a half-circle $C$ in the upper half complex plane with indentations at the poles, we have a residue coming from the pole in the upper-half complex plane $$ \oint_{C}\frac{dz}{1-z^4}=i2\pi\frac{1}{4i}=\frac{\pi}{2} $$ expanding the integration contour, neglecting the contribution from the large circle and evaluating the contribution from the poles on the real line, which in fact vanishes, $$ PV\int_{-\infty}^{+\infty}\frac{dt}{1-t^4}-i\pi\frac{1}{3}+i\pi\frac{1}{3}=\frac{\pi}{2}. $$ So $$ PV\int_{0}^{+\infty}\frac{dt}{1-t^4}=\frac{\pi}{4}, $$

Brightsun
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  • Thanks by the answer, thanks! but I am searching another way to solve the integral, by special function (beta and gamma) – Cheter Jan 17 '16 at 16:30
  • @Cheter You are right, sorry I misread the question slightly. – Brightsun Jan 17 '16 at 16:32
  • @Cheter I think I have something, see above! – Brightsun Jan 17 '16 at 16:59
  • Thanks for all, I want give you point but it do not allow me, thanks – Cheter Jan 17 '16 at 17:28
  • @Cheter No problem, however if you want add the final steps to my answer, you are welcome! – Brightsun Jan 17 '16 at 17:34
  • I'm just spitballing here, but what about splitting the integral into two integrals, one from 0 to 1, and 1 from 1 to inf. Then use the substitution t = 1/x on the second integral. After recombining I get the integral from 0 to 1 of x^2 and 1 / (1 - x^2)., and I think I dodged the convergence problems at x = 1. Also, principal value – Emanuel Landeholm Apr 15 '23 at 15:41
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Why special functions ?

$$\int_0^\infty\frac{dt}{1-t^4}=\int_0^\infty\left(\frac1{4(1-t)}+\frac1{4(1+t)}+\frac1{2(1+t^2)}\right)dt =\left.\left(\frac14\ln\left|\frac{1+t}{1-t}\right|+\frac12\arctan t\right)\right|_0^\infty=\frac\pi4.$$

  • "Why special functions ?" Probably because the integrals $$\int_0^\infty\frac1{1-t}dt$$ and $$\int_0^\infty\frac1{1+t}dt$$ both diverge, as does the integral $$\int_0^\infty\left(\frac1{4(1-t)}+\frac1{4(1+t)}\right)dt.$$ – Did Jul 01 '16 at 20:51
  • What if we split the integral into an LHS and an RHS for x = 1, and then use the substitution t = 1/x for the RHS? I get something manageable here after recombining and some algebra. I think I'm dodging the problems with convergence, – Emanuel Landeholm Apr 15 '23 at 15:44
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[2]{\,\mathrm{Li}_{#1}\left(\,{#2}\,\right)} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\mathrm{P.V.}\int_{0}^{\infty}{\dd t \over 1 - t^{4}}} &\ \stackrel{\mathrm{def.}}{=}\ \lim_{\epsilon \to 0^{+}}\pars{% \int_{0}^{1 - \epsilon}{\dd t \over 1 - t^{4}} + \int_{1 + \epsilon}^{\infty}{\dd t \over 1 - t^{4}}} \\[3mm] & = \lim_{\epsilon \to 0^{+}}\pars{% \int_{0}^{1 - \epsilon}{\dd t \over 1 - t^{4}} + \int_{1/\pars{1 + \epsilon}}^{0}{t^{2} \over 1 - t^{4}}\,\dd t} \\[3mm] & = \lim_{\epsilon \to 0^{+}}\pars{% \int_{0}^{1 - \epsilon}{\dd t \over 1 - t^{4}} - \int_{0}^{1/\pars{1 + \epsilon}}{t^{2} - 1 \over 1 - t^{4}}\,\dd t - \int_{0}^{1/\pars{1 + \epsilon}}{\dd t \over 1 - t^{4}}} \\[3mm] & = \lim_{\epsilon \to 0^{+}}\pars{% -\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}}{\dd t \over 1 - t^{4}} + \int_{0}^{1/\pars{1 + \epsilon}}{\dd t \over 1 + t^{2}}}\tag{1} \end{align}

However, \begin{align} 0 &< \verts{-\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}}{\dd t \over 1 - t^{4}}} < \verts{\pars{{1 \over 1 + \epsilon} - 1 + \epsilon} {1 \over 1 - 1/\pars{1 +\epsilon}^{4}}} \\[3mm] & = \verts{-1 + \epsilon + {1 \over 2 + \epsilon} + {1 \over 2 + \epsilon\pars{2 + \epsilon}}}\ \to\ \stackrel{\epsilon\ \to\ 0}{\large\color{#f00}{0}}\tag{2} \end{align}
$$ \mbox{Then, with}\ \pars{1}\ \mbox{and}\ \pars{2},\quad \color{#f00}{\mathrm{P.V.}\int_{0}^{\infty}{\dd t \over 1 - t^{4}}} = \int_{0}^{1}{\dd t \over 1 + t^{2}} = \color{#f00}{\pi \over 4} $$
Felix Marin
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