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How can I evaluate

$$\int_1^\infty \dfrac {1}{1+x^4}\mathrm dx$$

Thanks in advance!!

jimjim
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J.H.
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    There is a standard procedure for evaluating integrals of rational functions. What problems are you having? – Karolis Juodelė Sep 14 '13 at 11:56
  • The antiderivatives for this are $$x\mapsto \dfrac 1 4\left(4\sqrt 2\arctan (\sqrt 2x + 1)+\dfrac 1{\sqrt 2} \log\left(\dfrac{x^2-\sqrt 2 x +1}{x^2+\sqrt 2 x +1}\right)\right)+C$$ so I doubt that's the way to go about this. – Git Gud Sep 14 '13 at 12:00
  • I think something has to be substituted, but because of my lack of experience, I can't it....Could you please help me?? – J.H. Sep 14 '13 at 12:02
  • The Maple command $$with(Student[Calculus1]): IntTutor(1/(x^4+1), x) $$ finds the antiderivative step by step with explanations. – user64494 Sep 14 '13 at 12:51
  • You should answer more clearly just what your problem is. Do you mean you cannot find the antiderivative because you cannot find a good substitution, or you have the antiderivative and you cannot find the limit as the upper bound goes to infinity, or something else? – Rory Daulton Dec 21 '15 at 23:52

4 Answers4

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Hints:

$$ x^4+1=(x^2+\sqrt2\,x+1)(x^2-\sqrt2\,x+1)$$

Since $\,x>0\;$ , we thus have

$$\int\frac{dx}{x^4+1}=\frac1{2\sqrt2}\left(\text{arctanh}\frac{\sqrt 2\,x}{x^2+1}-\arctan(1-\sqrt2\,x)+\arctan(1+\sqrt2\,x)\right)+C$$

DonAntonio
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$$2\int\frac1{1+x^4}dx=\int\frac{1+x^2}{1+x^4}dx-\int\frac{x^2-1}{1+x^4}dx$$

$$=\int\frac{1+\frac1{x^2}}{x^2+\frac1{x^2}}dx-\int\frac{1-\frac1{x^2}}{x^2+\frac1{x^2}}dx=\int\frac{1+\frac1{x^2}}{\left(x-\frac1x\right)^2+2}dx-\int\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}dx$$

Put $x-\frac1x=u$ in $$\int_1^\infty\frac{1+\frac1{x^2}}{\left(x-\frac1x\right)^2+2}dx$$ so the limit ranges from $0,\infty$

Put $x+\frac1x=v$ in $$\int_1^\infty\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}dx$$ so the limit ranges from $2,\infty$

lab bhattacharjee
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The integral we want is in purple.

$$\int_0^\infty \frac{\text{d}x}{1+x^4}=\int_0^1 \frac{\text{d}{x}}{1+x^4}+\color\purple{\int_1^{\infty}\frac{\text{d}x}{1+x^4}}$$

Based on the property:

$$\int_0^\infty \frac{\text{d}x}{1+x^n} =\frac{\pi}{n}\csc \frac{\pi}{n}$$

(Verified here)

$$\implies \frac{\pi}{4}\csc \frac{\pi}{4}=\color\red{\int_0^1 \frac{\text{d}{x}}{1+x^4}}+\color\purple{\int_1^{\infty}\frac{\text{d}x}{1+x^4}}$$

Using @DonAntonio's answer, use the Fundamental Theorem of Calculus to evaluate what is in red. That leads to a very easy solution to your integral, without relying on limits.

Community
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} &\color{#0000ff}{\large\int_{1}^{\infty}{\dd x \over 1 + x^{4}}} = \int^{1}_{0}{x^{2} \over 1 + x^{4}}\,\dd x = \sum_{n = 0}^{\infty}\pars{-1}^{n}\int_{0}^{1}x^{4n + 2}\,\dd x = \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over 4n + 3} \\[3mm]&= \sum_{n = 0}^{\infty}\pars{{1 \over 8n + 3} - {1 \over 8n + 7}} = {1 \over 16}\sum_{n = 0}^{\infty}{1 \over \pars{n + 3/8}\pars{n + 7/8}} = {1 \over 16}\,{\Psi\pars{7/8} - \Psi\pars{3/8} \over 7/8 - 3/8} \\[3mm]&= \color{#0000ff}{\large{1 \over 8}% \bracks{\Psi\pars{7 \over 8} - \Psi\pars{3 \over 8}} \approx 0.2437} \end{align} $\Psi\pars{z}$ is the ${\it\mbox{digamma function}}$.

Felix Marin
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