How could one solve $\int_{0}^{\infty} \frac{1}{1-t^4}dt$ with special functions? 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?
 A: (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
$$
PV\int_{0}^{+\infty}\frac{dt}{1-t^4}=\frac{\pi}{4},
$$
A: Why special functions ?
$$\int_0^\infty\frac{dt}{1-t^4}=\int_0^\infty\left(\frac1{4(1-t)}+\frac1{4(1+t)}+\frac1{2(1+t^2)}\right)dt
=\left.\left(\frac14\ln\left|\frac{1+t}{1-t}\right|+\frac12\arctan t\right)\right|_0^\infty=\frac\pi4.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\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}

$$
\mbox{Then, with}\ \pars{1}\ \mbox{and}\ \pars{2},\quad
\color{#f00}{\mathrm{P.V.}\int_{0}^{\infty}{\dd t \over 1 - t^{4}}} =
\int_{0}^{1}{\dd t \over 1 + t^{2}} = \color{#f00}{\pi \over 4}
$$
