How one would evaluate the following integral?

$$\int_{0}^{\infty}\frac{\log^{10}(x)}{1+x^3} \, \mathrm{d}x$$

I have tried substitution with no success as well as differentiation under integral sign. Can anyone help me please. I prefer not to use contour integration.

  • $\begingroup$ $t \equiv {1 \over 1 + x^{3}}$ should lead somehow to the use of the Beta function. $\endgroup$ Jul 11, 2016 at 3:27

3 Answers 3


Hint. One may use the Euler beta integral in the form $$ \int_0^\infty\frac{x^{-s}}{1+x^3}dx=\frac13\int_0^\infty\frac{u^{-(s+2)/3}}{1+u}du=\frac{\pi}{3\sin (\pi (s+2)/3)}, \quad -2<s<1, $$ giving, by differentiating ten times and putting $s=0$,

$$ \int_0^\infty\frac{(\log x)^{10}}{1+x^3}dx=\left.\frac{d^{10}}{ds^{10}}\left(\frac{\pi}{3\sin (\pi (s+2)/3)} \right)\right|_{s=0} =\frac{3786350\: \pi ^{11}}{177147 \sqrt{3}}. $$

  • $\begingroup$ @OlivierOloa.Sorry I posted wrong problem please see the edit. $\endgroup$
    – Apollo15
    Jul 10, 2016 at 10:34
  • $\begingroup$ What's t in the yellow blanket? $\endgroup$
    – Zau
    Jul 10, 2016 at 11:36
  • $\begingroup$ @OlivierOloa have you calculated it manually $\endgroup$
    – Apollo15
    Jul 10, 2016 at 11:46
  • 1
    $\begingroup$ @shikharsinghal No. $\endgroup$ Jul 10, 2016 at 11:50
  • $\begingroup$ I just wanted an another method for it as I think no one will be willing to calculate it manually.well thanks. $\endgroup$
    – Apollo15
    Jul 10, 2016 at 11:53

Caveat. I wrote the following answer not having seen the request by the OP that he would prefer not to use contour integration. Perhaps what follows can help make the case for and showcase contour integration for this integral which belongs to a class that has frequently appeared here at MSE.

Observe that if we are allowed to use a CAS (which would appear necessary for this problem) then we may compute

$$Q_n = \int_0^\infty \frac{\log^n x}{x^3+1}\; dx = \int_0^\infty f_n(x) \; dx$$


$$f_n(z) = \frac{\log^n z}{z^3+1}$$

by computing all $Q_n$ recursively by integrating $f_{n+1}(z), f_n(z), \ldots$ and so on around a keyhole contour with the slot on the positive real axis and the branch cut of the logarithm on that axis as well, with argument from $0$ to $2\pi.$ The poles of $f_n(z)$ are at $\rho_k = \exp(\pi i/3 + 2\pi ik /3)$ with $k=0,1,2.$ We obtain for the residues

$$\mathrm{Res}_{z=\rho_k} f_n(z) = \mathrm{Res}_{z=\rho_k} \frac{\log^n z}{z^3+1} \\ = \left. \frac{\log^n z}{3z^2} \right|_{z=\rho_k} = \left. z \frac{\log^n z}{3z^3} \right|_{z=\rho_k} = - \left. \frac{1}{3} z \log^n z \right|_{z=\rho_k} \\ = - \frac{1}{3} \exp(\pi i/3 + 2\pi ik /3) (\pi i/3 + 2\pi i k/3)^n = \alpha_{n,k}.$$

We obtain by integrating $f_n(z)$

$$\int_0^\infty \frac{\log^n z}{z^3+1} \; dz + \int_\infty^0 \frac{(2\pi i + \log z)^n}{z^3+1} \; dz \\ = 2\pi i \sum_k \mathrm{Res}_{z=\rho_k} f_n(z) = 2\pi i \sum_k \alpha_{n,k}.$$

This yields

$$ \sum_{p=0}^{n-1} {n\choose p} (2\pi i)^{n-p} \int_0^\infty \frac{\log^p z}{z^3+1} \; dz = - 2\pi i \sum_k \alpha_{n,k}$$

which is

$$ \sum_{p=0}^{n-1} {n\choose p} (2\pi i)^{n-p-1} Q_p = - \sum_k \alpha_{n,k}$$


$$\sum_{p=0}^{n} {n+1\choose p} (2\pi i)^{n-p} Q_p = - \sum_k \alpha_{n+1,k}$$

Therefore to compute $Q_n$ we use the recurrence

$$Q_n = - \frac{1}{n+1} \sum_k \alpha_{n+1,k} - \frac{1}{n+1} \sum_{p=0}^{n-1} {n+1\choose p} (2\pi i)^{n-p} Q_p$$

We just need the base case $Q_0$ which we compute using a pizza slice resting on the positive real axis and having argument $2\pi/3$ so that it only contains $\alpha_{0,0}.$ Parameterizing with $z=\exp(2\pi i/3) t$ we get

$$Q_0 - \exp(2\pi i/3) Q_0 = 2\pi i \alpha_{0,0} $$

which yields

$$Q_0 = - \frac{1}{3} 2\pi i \frac{\exp(\pi i/3)}{1-\exp(2\pi i/3)} = - \frac{1}{3} 2\pi i \frac{1}{\exp(-\pi i/3)-\exp(\pi i/3)} \\ = \frac{1}{3} \pi \frac{1}{\sin(\pi/3)} = \frac{2}{9}\sqrt{3}\pi.$$

With everything in place we obtain e.g. the sequence up to $n=10$

$$-{\frac {2\,{\pi }^{2}}{27}},{\frac {10\,{\pi }^{3} \sqrt {3}}{243}},-{\frac {14\,{\pi }^{4}}{243}},{\frac {34\,{\pi }^{5}\sqrt {3}}{729}},\\-{\frac {806\,{\pi }^{6 }}{6561}},{\frac {910\,{\pi }^{7}\sqrt {3}}{6561}},-{ \frac {10414\,{\pi }^{8}}{19683}},{\frac {415826\,{\pi }^{9}\sqrt {3}}{531441}},\\-{\frac {685762\,{\pi }^{10}}{ 177147}},{\frac {3786350\,{\pi }^{11}\sqrt {3}}{531441}}, \ldots$$

The Maple code for this is extremely simple, consisting of a few lines.

alpha := (n,k) ->
-1/3 * exp(Pi*I/3+2*Pi*I*k/3) * (Pi*I/3 + 2*Pi*I*k/3)^n;

Q :=
option remember;
    local res;

    if n = 0 then return 2/9*sqrt(3)*Pi fi;

    res :=
    -1/(n+1)*add(alpha(n+1,k), k=0..2)
    -1/(n+1)*add(binomial(n+1, p)*(2*Pi*I)^(n-p)*Q(p),


VERIF := n -> int((log(x))^n/(x^3+1), x=0..infinity);

Observe that this method generalizes quite nicely. Suppose we are interested in

$$K_n = \int_0^\infty \frac{\log^n x}{x^3-2x+4} \; dx.$$

The same computation goes through except now we have the following three poles and their logarithms

$$\begin{array}{|l|l|} \hline \text{pole} & \text{logarithm} \\ \hline \rho_0 = 1+i & \log \rho_0 = \frac{1}{2} \log 2 + \frac{1}{4}i\pi \\ \hline \rho_1 = 1-i & \log \rho_1 = \frac{1}{2} \log 2 + \frac{7}{4}i\pi \\ \hline \rho_2 = -2 & \log \rho_2 = \log 2 + i\pi.\\ \hline \end{array}$$

The rest is unchanged. We obtain e.g.

$$K_4 = {\frac {357\,{\pi }^{5}}{10240}}-{\frac {31\, \left( \ln \left( 2 \right) \right) ^{5}}{1600}}-{\frac {139\, \left( \ln \left( 2 \right) \right) ^{3}{\pi }^{2}}{1920 }} \\ -{\frac {4897\,\ln \left( 2 \right) {\pi }^{4}}{76800}}+ {\frac {9\, \left( \ln \left( 2 \right) \right) ^{4}\pi }{640}}+{\frac {63\, \left( \ln \left( 2 \right) \right) ^{2}{\pi }^{3}}{1280}}.$$

The Maple code is very similar to the first version.

alpha_sum :=
local poles;

    poles :=
    [[1+I, 1/2*log(2) + I*Pi/4],
     [1-I, 1/2*log(2) + 7*I*Pi/4],
     [-2, log(2) + I*Pi]];

    add(residue(1/(x^3-2*x+4), x=p[1])*p[2]^n,
                p in poles);

Q :=
option remember;
    local res;

    if n = 0 then
        simplify(int(1/(x^3-2*x+4), x=0..infinity));

    res :=
    -1/(n+1)*add(binomial(n+1, p)*(2*Pi*I)^(n-p)*Q(p),


VERIF := n -> int((log(x))^n/(x^3-2*x+4), x=0..infinity);
  • $\begingroup$ it is excellent $\endgroup$
    – Apollo15
    Jul 15, 2016 at 11:50
  • $\begingroup$ Marko, I was endeavoring to proceed along precisely this way forward to developing a solution for a similar post. But, I saw your comment on that post, had a look, and decided I would defer. +1 for this thorough answer! -Mark $\endgroup$
    – Mark Viola
    Oct 20, 2016 at 0:38
  • $\begingroup$ Thank you for these kind words. The post where I added the comment just now requires a bit more, asking for an EGF of the values, which I might look at tomorrow, feel free to solve this part if you are interested. $\endgroup$ Oct 20, 2016 at 0:41
  • $\begingroup$ @Dr.MV The answer to this part is at the following MSE link. $\endgroup$ Oct 21, 2016 at 0:34

$n\geq 2,m\geq 0$, integer,

\begin{align}J_n(m)=\int_0^\infty \frac{\ln^m x}{1+x^n}\,dx\end{align}

Observe that,

\begin{align}J_n(m)&=\int_0^1 \frac{\ln^m x}{1+x^n}\,dx+\int_1^\infty \frac{\ln^m x}{1+x^n}\,dx\\ \end{align}

In the latter integral perform the change of integral $y=\dfrac{1}{x}$,

\begin{align}J_n(m)&=\int_0^1 \frac{(1+(-1)^mx^{n-2})\ln^m x}{1+x^n}\,dx\end{align}

Observe that for $x \in [0;1[$,

\begin{align} \frac{1-x}{1+x^3}=\frac{1}{1-x}-\frac{2x}{1-x^2}+\frac{2x^5}{1-x^6}-\frac{x^2}{1-x^3}\end{align}


\begin{align}J_3(2m+1)&=\int_0^1 \frac{\ln^{2m+1} x}{1-x}\,dx-\int_0^1 \frac{2x\ln^{2m+1} x}{1-x^2}\,dx+\int_0^1 \frac{2x^5\ln^{2m+1} x}{1-x^6}\,dx-\int_0^1 \frac{x^2\ln^{2m+1} x}{1-x^3}\,dx\\\end{align}

In the second integral perform the change of variable $y=x^2$, In the third integral perform the change of variable $y=x^6$, In the fourth integral perform the change of variable $y=x^3$,

\begin{align}J_3(2m+1)&=\left(1-\frac{1}{2^{2m+1}}+\frac{1}{3\times 6^{2m+1}}-\frac{1}{3^{2m+2}}\right)\int_0^1\frac{\ln^{2m+1} x}{1-x}\,dx\\ &=(2m+1)!\left(\frac{1}{2^{2m+1}}-\frac{1}{3\times 6^{2m+1}}+\frac{1}{3^{2m+2}}-1\right)\zeta(2m+2) \end{align}


\begin{align}K_n(m)&=\int_0^\infty \int_0^\infty \frac{\ln^m(xy)}{(1+x^n)(1+y^n)}\,dx\,dy\\ &=\sum_{k=0}^{m}\binom{m}{k}J_n(k)J_n(m-k)\\ \end{align}

On the other hand, perform the change of variable $u=yx$,

\begin{align}K_n(m)&=\int_0^\infty \int_0^\infty \frac{y^{n-1}\ln^m(u)}{(u^n+y^n)(1+y^n)}\,du\,dy\\ \end{align}

perform the change of variable $v=y^n$,

\begin{align}K_n(m)&=\frac{1}{n}\int_0^\infty \int_0^\infty \frac{\ln^m(u)}{(u^n+v)(1+v)}\,du\,dv\\ &=\frac{1}{n}\int_0^\infty \left[\frac{1}{u^n-1}\ln\left(\frac{v+1}{v+u^n}\right)\right]_0^\infty \ln^m(u)\,du\\ &=\int_0^\infty \frac{\ln^{m+1}(u)}{u^n-1}\,du\\ \end{align}

\begin{align}K_3(2m)&=\int_0^\infty \frac{\ln^{2m+1}(u)}{u^3-1}\,du\\ &=\int_0^1 \frac{(1+u)\ln^{2m+1}(u)}{u^3-1}\,du\\ &=\int_0^1 \frac{\ln^{2m+1}(u)}{u-1}\,du-\int_0^1 \frac{u^2\ln^{2m+1}(u)}{u^3-1}\,du \end{align}

In the latter integral perform the change of variable $y=u^3$,

\begin{align}K_3(2m)&=\left(1-\frac{1}{3^{2m+2}}\right)\int_0^1 \frac{\ln^{2m+1}(u)}{u-1}\,du\\ &=(2m+1)!\left(1-\frac{1}{3^{2m+2}}\right)\zeta(2m+2) \end{align}


\begin{align}\sum_{k=0}^{2m}\binom{2m}{k}J_3(k)J_3(2m-k)=(2m+1)!\left(1-\frac{1}{3^{2m+2}}\right)\zeta(2m+2) \end{align}

Observe that,




If you know that,




Thus, if $m>0$,

\begin{align}J_3(2m)=\frac{3\sqrt{3}}{4\pi}\left((2m+1)!\left(1-\frac{1}{3^{2m+2}}\right)\zeta(2m+2)-\sum_{k=1}^{2m-1}\binom{2m}{k}J_3(k)J_3(2m-k)\right) \end{align}

If you know $\zeta(2),...,\zeta(2m+2)$ then you can compute $J_3(2m)$

Here is code for GP PARI:



Code assumes that:

\begin{align}\int_0^\infty \frac{\ln^{2n}x}{1+x^3}\,dx=\frac{a_n}{b_n}\sqrt{3}\pi^{2n+1}\end{align} $a_n,b_n$ natural integers.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .