Find this integral $$I=\int_{0}^{\infty}\dfrac{\ln^3{x}}{(1+x^2)(1+x)^2}dx$$

My try: let $x=\tan{t}$ then $$I=\int_{0}^{\frac{\pi}{2}}\dfrac{\ln^3{\tan{t}}}{(1+\tan{t})^2}dt$$ I am unable to simplify after this. This problem is from QQ.

  • 1
    Have you tried complex integration using a certain contour? – user88595 Dec 17 '13 at 14:09
  • Where did this integral come from? – Mhenni Benghorbal Dec 17 '13 at 14:15
  • A month ago, I asked the question : is there a limit to teachers and textbooks imagination ? I never got any answer but obviously, if the limit exist, it is somewhere outside the universe. – Claude Leibovici Dec 17 '13 at 14:35
  • Just curious. What does QQ mean? – user17762 Dec 17 '13 at 22:30
up vote 16 down vote accepted

Here is an elementary way. First note that $$\int_1^{\infty} \dfrac{\ln^3(x) dx}{(1+x^2)(1+x)^2} = \int_1^0 \dfrac{-\ln^3(x)}{(1+1/x^2)(1+1/x)^2} \dfrac{-dx}{x^2} = \int_0^1 \dfrac{-x^2 \ln^3(x)}{(1+x^2)(1+x)^2}dx$$ Hence, your integral is $$I = \int_0^1 \dfrac{(1-x^2) \ln^3(x)}{(1+x^2)(1+x)^2}dx = \underbrace{\int_0^1 \dfrac{\ln^3(x)}{1+x}dx}_J - \overbrace{\int_0^1 \dfrac{x\ln^3(x)}{1+x^2}dx}^K$$ We have $$J = \sum_{k=0}^{\infty}(-1)^k\int_0^1 x^k \ln^3(x)dx = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac6{(k+1)^4} = -\dfrac7{120} \pi^4 \tag{$\star$}$$ We have $$K = \sum_{k=0}^{\infty}(-1)^k\int_0^1 x^{2k+1} \ln^3(x)dx = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac3{8(k+1)^4} = -\dfrac7{1920} \pi^4 \tag{$\dagger$}$$ Hence, $$\boxed{\color{red}{I = J-K = -\dfrac7{128}\pi^4}}$$ Where we used the following facts $$\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(k+1)^4} = -\dfrac78 \zeta(4)$$ $$\zeta(4) = \dfrac{\pi^4}{90}$$ to simplify $(\star)$ and $(\dagger)$.

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    I really like this method +1 =) – N3buchadnezzar Dec 18 '13 at 0:40

You can use the residue theorem. Consider the integral

$$\oint_C dz \frac{\log^4{z}}{(1+z^2)(1+z)^2}$$

where $C$ is a keyhole contour about the positive real axis, so that $\arg{z} \in [0,2 \pi)$. $C$ has an outer radius of $R$, and an inner radius of $\epsilon$. The magnitude of the integral vanishes along the outer arc as $2 \pi \log^4{R}/R^3$ as $R \to \infty$ and along the inner arc as $\epsilon \log^4{\epsilon}$ as $\epsilon \to 0$. Thus the contour integral is equal to, in these limits

$$\int_0^{\infty} dx \frac{\log^4{x}-(\log{x}+i 2 \pi)^4}{(1+x^2)(1+x)^2}$$

which, when expanded, is equal to

$$-i 8 \pi \int_0^{\infty} dx \frac{\log^3{x}}{(1+x^2)(1+x)^2}+ 24 \pi^2 \int_0^{\infty} dx \frac{\log^2{x}}{(1+x^2)(1+x)^2}\\+i 32 \pi^3 \int_0^{\infty} dx \frac{\log{x}}{(1+x^2)(1+x)^2}-16 \pi^4 \int_0^{\infty} dx \frac{1}{(1+x^2)(1+x)^2}$$

The contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles $z_{1,2}=\pm i$ and $z_3=-1$. Note that we would then have to evaluate the integrals with lower powers of log. We may circumvent this by expressing the above equation as a system of equations for the unknown integrals. Let

$$R_j = \sum_{k=1}^3 \operatorname*{Res}_{z=z_k} \frac{\log^j{z}}{(1+z^2)(1+z)^2}$$

$$I_j = \int_0^{\infty} dx \frac{\log^j{x}}{(1+x^2)(1+x)^2}$$

Thus, by considering similar contour integrals in the complex plane, we have the following system of equations:

$$\begin{align}-i 8 \pi I_3+24 \pi^2 I_2+i 32 \pi^3 I_1-16 \pi^4 I_0 &= i 2 \pi R_4\\ -i 6 \pi I_2+12 \pi^2 I_1+i 8 \pi^3 I_0&=i 2 \pi R_3\\-i 4 \pi I_1+4 \pi^2 I_0 &= i 2 \pi R_2\\-i 2 \pi I_0 &= i 2 \pi R_1\end{align} $$

We may now solve this upper-diagonal system for the integrals in terms of the residues; we are only interested in $I_3$. Solving for $I_3$ and reexpressing in terms of the original notation, we find that our integral is

$$\int_0^{\infty} dx \frac{\log^3{x}}{(1+x^2)(1+x)^2} = \sum_{k=1}^3 \operatorname*{Res}_{z=z_k} \left [\frac{-\frac14 \log^4{z}+i \pi \log^3{z}+\pi^2 \log^2{z}}{(1+z^2)(1+z)^2} \right ]$$

Now we must evaluate the residues. At the poles $z_1=e^{i \pi/2}$ and $z_2=e^{i 3 \pi/2}$, the computation is straightforward:

$$\operatorname*{Res}_{z=e^{i \pi/2}} \left [\frac{-\frac14 \log^4{z}+i \pi \log^3{z}+\pi^2 \log^2{z}}{(1+z^2)(1+z)^2} \right ] = \\ \frac{-\frac14 (i \pi/2)^4+i \pi (i \pi/2)^3+\pi^2 (i \pi/2)^2}{2 i (1+i)^2}=\frac{9\pi^4}{256}$$

Similarly,

$$\operatorname*{Res}_{z=e^{i 3\pi/2}} \left [\frac{-\frac14 \log^4{z}+i \pi \log^3{z}+\pi^2 \log^2{z}}{(1+z^2)(1+z)^2} \right ] = \frac{9\pi^4}{256}$$

For the pole at $z=e^{i \pi}$, we must differentiate to evaluate the residue (double pole). Thus,

$$\operatorname*{Res}_{z=e^{i \pi}} \left [\frac{-\frac14 \log^4{z}+i \pi \log^3{z}+\pi^2 \log^2{z}}{(1+z^2)(1+z)^2} \right ] = \left [\frac{d}{dz} \frac{-\frac14 \log^4{z}+i \pi \log^3{z}+\pi^2 \log^2{z}}{1+z^2} \right ]_{z=e^{i \pi}}$$

which calculation I will spare you at this point, except to say that it is straightforward, and has remarkable cancellation of the imaginary part. The result of this calculation informs us that the residue is $-\pi^4/8$.

Finally, putting this all together, we have

$$\int_0^{\infty} dx \frac{\log^3{x}}{(1+x^2)(1+x)^2} = \frac{9 \pi^4}{256} + \frac{9\pi^4}{256}-\frac{\pi^4}{8} = -\frac{7 \pi^4}{128}$$

ADDENDUM

It should be understood that the expression for the integral in terms of residues is not specific to this particular integral and applies to any integral of the form

$$\int_0^{\infty} dx \, f(x) \, \log^3{x} = \sum_{k=1}^N \operatorname*{Res}_{z=z_k} \left [\left (-\frac14 \log^4{z}+i \pi \log^3{z}+\pi^2 \log^2{z}\right ) f(z) \right ]$$

where $f$ is sufficiently well behaved that the integral exists, and the $z_k$ are the poles of $f$ in the complex plane away from the positive real axis. In fact, the general procedure works for any integer power of log, and it would be interesting to generate a polynomial-type expression in log for arbitrary powers.

  • Just curious, in what sort of book would I first look to learn this sort of integration? Is it in a complex analysis book, or something else? – apnorton Dec 17 '13 at 21:27
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    @anorton: Sigh. The closest thing I can think of is a book called The Cauchy Method of Residues, which has all sorts of methods of attack using the residue theorem. The problem here, though, is that we have a different situation: standard contour, but really potentially painful evaluation due to cube of log. I think this methodology is something I developed while working the problem out myself. Likely not original, but no idea where else you could find this result. If you want something basic, Churchill & Brown is a good undergrad reference. – Ron Gordon Dec 17 '13 at 21:30
  • Wonderful post. – Marko Riedel Dec 18 '13 at 1:38
  • @RonGordon Just wanted to share a thought; Though complex analysis is a powerful tool, sometime methods based on real analysis, have their own charm. – user17762 Dec 18 '13 at 3:25
  • @RonGordon Also, thanks for suggesting the book "The Cauchy Method of Residues", which I was able to get hold of from my departmental library. Interestingly I have never heard of this book before. – user17762 Dec 18 '13 at 3:28

Here is a closed form

$$I=\int_{0}^{\infty}\dfrac{\ln^3{x}}{(1+x^2)(1+x)^2}dx = -\frac{7}{128} \pi^4 \sim -5.327059668.$$

You can use this technique.

  • Where did your answer come from? – Ron Gordon Dec 17 '13 at 14:34
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    Yes. Totally not obvious how you would apply that technique to this problem. Perhaps I lack imagination, but many of us do not have a table of Mellin transforms handy. – Ron Gordon Dec 17 '13 at 14:37
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    @ClaudeLeibovici: But doesn't that expression include several polylogarithms ? – Lucian Dec 17 '13 at 15:03
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    Mhenni, not to be harsh, but I could have stated the same thing about residue theory: "Here, use this technique, here's the answer." A little work shows that this is completely inadequate for providing any clue to the answer. – Ron Gordon Dec 17 '13 at 16:38
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    OK, I have a relevant question (or three): did you use the linked technique to derive the result? And if so, maybe you'd like to show a step or two. And if not, then how did you, and why was that way not good enough to advertise, and why did you not use the way you did advertise? – Ron Gordon Dec 23 '13 at 14:26

What I propose only aims at simplifying the evaluation. Consider the following contour integral: $$\oint_\gamma \frac{\log^4(z)}{(1+z^2)(1\color{red}{-}z)^2}dz=2\pi i \sum \text{Res}f(z)$$ Notice the minus in $\color{red}{\text{red}}$ in the denominator instead of the plus in the original function; you will soon find out why I did this. Taking $\gamma$ to be the keyhole contour about the negative axis, and setting $z=-x-i\epsilon$ for the lower line and $z=-x+i\epsilon$ for the upper, you get: $$-\int_\infty^0 \frac{\log^4(-x+i\epsilon)}{(1+(-x+i\epsilon)^2)(1\color{red}{-}(-x+i\epsilon))^2}dx-\int_0^\infty \frac{\log^4(-x-i\epsilon)}{(1+(-x-i\epsilon)^2)(1\color{red}{-}(-x-i\epsilon))^2}dx$$ $$\int_0^\infty \frac{(\log(x)+i\pi)^4}{(1+x^2)(1\color{blue}{+}x)^2}dx-\int_0^\infty \frac{(\log(x)-i\pi)^4}{(1+x^2)(1\color{blue}{+}x)^2}dx$$

$$\int_0^\infty\frac{8\pi i(\log^3(z)-\pi^2\log(z))}{(1+x^2)(1+x)^2}dx=8\pi i\left(I-\pi^2J\right)$$

$I$ is the integral that you're looking for, while $J$ is the integral containing $\log(z)$ which can be evaluated using the same trick (keyhole contour). It turns out to be $$J=-\frac{\pi^2}{16}$$

Furthermore, the residues in the first equation above can be found in a straightforward way, the sum turns out to be: $$\sum\text{Res}f(z)=\frac{\pi^4}{32}$$ so that: $$8\pi i\left(I+\frac{\pi^4}{16}\right)=2\pi i \frac{\pi^4}{32}$$ Solving for $I$ we get: $$\boxed{\color{blue}{I=-\frac{7}{128}\pi^4}}$$

  • There is one less term along this branch cut, so congrats. However, evaluating the lower-order integrals is a really interesting recursive process that leads to a lower-diagonal linear system. You just waved your hand over it like a magic elixir, so your solution is missing what I at least feel are the really cool parts. – Ron Gordon Aug 12 '16 at 15:48
  • @RonGordon I totally agree, I wasn't going to copy your method, my aim was to get a system with fewer equations in order to reduce the painful evaluation you mentioned in your comments. I prefer the negative axis for this reason, plus the function in the contour integral is slightly different from the one in the original integral, so that's a bit interesting, though obviously nowhere near as interesting as the recursion! Thank you for the feedback. – GeorgSaliba Aug 12 '16 at 15:55
  • I did not say that the evaluation was painful but interesting. I took the time to detail how to eliminate the evaluation of all of these separate integrals by instead solving a simple linear system and reducing the multiple integrals into a single residue calculation. It is the opposite of painful! I actually thought about expanding on this for arbitrary powers of log and writing up the results. If only I had the time... – Ron Gordon Aug 12 '16 at 19:56
  • @RonGordon I must have misunderstood your comment from three years ago. And I hope you do find the time... – GeorgSaliba Aug 12 '16 at 20:11

Filling in the steps of Mhenni Benghorbal's solution we have

The Mellin transform \begin{equation} \mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x \label{eq:160813a2} \tag{2} \end{equation}

where \begin{equation} f(x) = \frac{1}{(1+x^{2})(1+x)^{2}} = -\frac{1}{2}\frac{x}{x^{2}+1} + \frac{1}{2}\frac{1}{x+1} + \frac{1}{2}\frac{1}{(x+1)^{2}} \label{eq:160813a3} \tag{3} \end{equation} via partial fraction expansion.

Applying the Mellin transform, yields \begin{align} \mathcal{M}[f(x)](s) & = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(1+x^{2})(1+x)^{2}} \mathrm{d} x \\ & = -\frac{1}{2}\left[\frac{1}{2}\pi\sec\left(\frac{\pi}{2}s\right)\right] + \frac{1}{2}\pi\csc(\pi s) + \frac{1}{2} \mathrm{B}(s,2-s) \label{eq:160813a4} \tag{4} \end{align}

Taking the 3rd derivative of equation \eqref{eq:160813a4} with respect to s and then taking $\lim s \to 1$ yields \begin{align} \int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x & = -\frac{1}{2}\left( -\frac{7}{960} \pi^{4} \right) + \frac{1}{2} \left( -\frac{7}{60} \pi^{4} \right) + \frac{1}{2}0 \\ & = -\frac{7}{128} \pi^{4} \label{eq:160813a5} \tag{5} \end{align}

Let us fill in the details. Handling the beta function first, we have \begin{equation} \mathrm{B}(s,2-s) = \frac{\Gamma(s)\Gamma(2-s)}{\Gamma(2)} = \Gamma(s)\Gamma(2-s) \label{eq:160813a6} \tag{6} \end{equation} To take derivatives, we note that \begin{equation} \frac{\mathrm d}{\mathrm d s} \Gamma(s) = \Gamma(s) \psi^{(0)}(s) \quad \mathrm{and} \quad \psi^{(n)}(s) = \frac{\mathrm{d}^{n}}{\mathrm{d} s^{n}} \psi^{(0)}(s) \end{equation} Where $\psi^{(n)}(s)$ is the polygamma function.

Taking the third derivative of equation \eqref{eq:160813a6} and letting $\lim s \to 1$ equals 0. Here we used \begin{equation} \psi^{(0)}(1) = -\gamma \quad \mathrm{and} \quad \psi^{(1)}(1) = \frac{\pi^{2}}{6} \end{equation} and fortunately $\Gamma^{(3)}(s) = -\Gamma^{(3)}(2-s)$ which leads to some cancellations.

Doing the same for the first two terms on the right hand side of equation \eqref{eq:160813a4}, we have to be careful. Each of them individually goes to $\infty$ as $\lim s \to 1$ but the $(s-1)^{-4}$ terms in the Laurent expansions about $s=1$ cancel. Here I used Wolfram Alpha. Doing it with the two terms combined yielded our final answer.

I think what Mhenni Benghorbal suggested is this: \begin{align*} I\left(a\right): & =\int_{0}^{\infty}\frac{x^{a}}{\left(1+x^{2}\right)\left(1+x\right)^{2}}dx=\frac{\pi}{\sin\pi a}\left(\frac{1}{2}\cos\frac{\pi a}{2}+\frac{a-1}{2}\right). \end{align*} Then \begin{align*} \left(\frac{d}{da}\right)^{m}I\left(0\right) & =\int_{0}^{\infty}\frac{\left(\ln x\right)^{m}}{\left(1+x^{2}\right)\left(1+x\right)^{2}}dx\\ & =\left(\frac{d}{da}\right)^{m}\Big|_{a=0}\,\,\frac{\pi}{\sin\pi a}\left(\frac{1}{2}\cos\frac{\pi a}{2}+\frac{a-1}{2}\right). \end{align*}

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