By reading a german wikipedia (see here) about integrals, i stumpled upon this entry

27 1.5 $$ \color{black}{ \int_0^\infty \frac{1}{x+1-u}\cdot \frac{\mathrm{d}x}{\log^2 x+\pi^2} =\frac{1}{u}+\frac{1}{\log(1-u)}\,, \qquad u \in (0,1)} $$

(Click for the source) Where the result was proven using complex analysis. Is there any method to show the equality using real methods? Any help will be appreciated =)

  • $\begingroup$ Do you mean by $u\in\mathbb{R}^{\ge1}$ that $u\ge1$? If that is the case, the integrand is not integrable near $x=u-1$. Do you want the Cauchy Principal Value? Furthermore, if $u\gt1$, $\log(1-u)$ is complex, not good for real evaluation. $\endgroup$
    – robjohn
    Aug 12, 2013 at 21:08
  • $\begingroup$ The source says that $u\not\in\mathbb{R}^{\ge1}$. $\endgroup$
    – robjohn
    Aug 12, 2013 at 21:13
  • $\begingroup$ I think a substitution of $y=\log{x}$ may be helpful here, but I haven't taken it all the way yet. $\endgroup$
    – Ron Gordon
    Aug 12, 2013 at 23:18
  • $\begingroup$ Norberts answer here math.stackexchange.com/questions/208345/… seems promising. But adding the $u$ does not give the wanted cancellation. $\endgroup$ Aug 12, 2013 at 23:40
  • $\begingroup$ The real line approach is usually to find an anti-derivative and evaluate it at the end points. If you can't find an anti-derivative (and even if you can and it's a mess) using a complex contour integral may solve your problem, as it appears to have done here. If you are determined to avoid complex functions, you might try building more easily integrable successive approximations to your integrand; then show that the limit of the integral of the approximations is the integral of the limit. You could look at the Taylor's series, although they don't converge uniformly. $\endgroup$
    – Betty Mock
    Aug 13, 2013 at 23:27

3 Answers 3


By setting $e^\eta=v=1-u$ and exploiting the inverse Laplace transform we have:

$$\int_{1}^{+\infty}\frac{dx}{(x+v)\left(\pi^2+\log^2 x\right)}=\frac{1}{\pi}\int_{1}^{+\infty}\frac{dx}{x+v}\int_{0}^{+\infty}\sin(\pi z)\,x^{-z}\,dz.\tag{1}$$ Moreover, if $0<z<1$, by exploiting the Euler beta function and the reflection formulas for the $\Gamma$ function we have: $$ \int_{0}^{+\infty}\frac{x^{-z}}{x+v}\,dx = \frac{\pi}{v^z\sin(\pi z)}=\int_{0}^{+\infty}\frac{x^z}{x+vx^2}\,dx\tag{2}$$ and rearranging carefully we get:

$$ \int_{0}^{+\infty}\frac{dx}{(x+v)\left(\pi^2+\log^2 x\right)}=\int_{0}^{1}t^{-v}\,dt - \int_{0}^{+\infty}v^{-z}\,dz = \frac{1}{1-v}-\frac{1}{\log v}\tag{3}$$ as wanted. Anyhow, this is not the sketch of a really alternative proof, since the inverse Laplace transform is just the residue theorem in disguise.


I'm not sure about the full solution, but there is a way to find an interesting functional equation for this integral.

First, let's get rid of the silly restriction on $u$. By numerical evaluation, the integral exists for all $u \in (-\infty,1)$

enter image description here

Now let's introduce the more convenient parameter:


$$I(v)=\int_0^{\infty} \frac{dx}{(v+x)(\pi^2+\ln^2 x)}$$

Now let's make a change of variable:


$$I(v)=\int_{-\infty}^{\infty} \frac{e^t dt}{(v+e^t)(\pi^2+t^2)}$$

$$I(v)=\int_{-\infty}^{\infty} \frac{(v+e^t) dt}{(v+e^t)(\pi^2+t^2)}-v \int_{-\infty}^{\infty} \frac{ dt}{(v+e^t)(\pi^2+t^2)}=1-v J(v)$$

Now let's make another change of variable:


$$I(v)=\int_{-\infty}^{\infty} \frac{e^{-z} d(-z)}{(v+e^{-z})(\pi^2+z^2)}=\int_{-\infty}^{\infty} \frac{ dz}{(1+v e^z)(\pi^2+z^2)}=\frac{1}{v} J \left( \frac{1}{v} \right)$$

Now we get:

$$1-v J(v)=\frac{1}{v} J \left( \frac{1}{v} \right)=I(v)$$

$$v J(v)+\frac{1}{v} J \left( \frac{1}{v} \right)=1$$

$$v \in (0,\infty)$$

For example, we immediately get the correct value:

$$J(1)=I(1)=\int_0^{\infty} \frac{dx}{(1+x)(\pi^2+\ln^2 x)}=\frac{1}{2}$$

We can also check that this equation works for the known solution (which is actually valid on the whole interval $v \in (0,\infty)$, except for $v=1$).

$$I(v)=\frac{1}{1-v}+\frac{1}{\ln v}$$

enter image description here

$$J(v)=-\frac{1}{1-v}-\frac{1}{v \ln v}$$

$$J \left( \frac{1}{v} \right)=\frac{v}{1-v}+\frac{v}{\ln v}$$

$$1-v J(v)=\frac{1}{v} J \left( \frac{1}{v} \right)$$

Now this is not a solution of course (except for $I(1)$), but it's a big step made without any complicated integration techniques.

Basically, if we define:



We need to solve a simple functional equation:

$$I(v)+I \left( \frac{1}{v} \right)=1$$


Below is a real method by utilizing $\int_0^1 \sin(\pi t)x^{-t}dt=\frac{\pi(x+1)}{x(\ln^2 x+\pi^2)}$

\begin{align} I=&\int_0^\infty \frac{1}{(x+1-u)(\ln^2 x+\pi^2)} \ dx\\ =& \int_0^\infty \frac{1}{x+1-u}\bigg( \frac1\pi\int_0^1 \sin(\pi t)x^{-t}dt -\frac1{x(\ln^2 x+\pi^2)}\bigg)dx\\ =&\ \frac1\pi\int_0^1\sin(\pi t) \left(\int_0^\infty\frac{x^{-t}}{x+1-u}dx\right) dt\\ &\hspace{10mm}-\frac1{1-u}\int_0^\infty\left(\frac1x-\frac1{x+1-u}\right)\frac1{\ln^2 x+\pi^2}dx\\ =& \int_0^1 (1-u)^{-t}dt-\frac1{1-u}+\frac1{1-u}I\\ =& \ \frac{1-u}{-u}\left( \frac {-u}{(1-u)\ln(1-u)}-\frac1{1-u}\right)\\ =&\ \frac{1}{u}+\frac{1}{\ln(1-u)} \end{align} where $\int_0^\infty\frac{x^{-t}}{x+1-u}dx=\frac{\pi(1-u)^{-t}}{\sin(\pi t)}$ and $ \int_0^\infty \frac1{x(\ln^2 x+\pi^2)}dx= 1$ are used in above evaluation.


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