When evaluating definite and indefinite integrals, there are times in which the integrand presents itself to be solvable using a definite method: be it a transformation, substitution, series, etc.

For instance, the general method for integrals involving rational functions of trigonometric functions is to employ the Weirerstrauss Substitution.

Now, on this site I mainly primarily focus on integrals and I've observed that the substitution of $y = \frac{1 + x}{1 - x}$ is used when the bounds of a definite integral are $0,1$. Is this part of a generalised method? and if so, does it have name?

And if not, is it possible to get some examples of integrals where this method is favourable.

  • $\begingroup$ As omegadot already pointed out within his answer the substitution $y=\frac{1-x}{1+x}$ is more convenient when dealing within the interval $[0,1]$, and especially while dealing with the borders of integration $0$ and $1$ since there are mapped to themself. On the other hand applying your substituion produces the new bordes $0$ and $\infty$ which, indeed, can be useful under certain circumstances aswell. $\endgroup$
    – mrtaurho
    Mar 1 '19 at 10:43


1) As already pointed out $\phi\left([0;1]\right)=[0;1]$

2) the change of variable $y=\phi(x)$ preserves both:

\begin{align}&\frac{dx}{1+x^2}\\ &\frac{dx}{1+x} \end{align}

3) \begin{align}1+\phi(x)&=\frac{2}{1+x}\\ 1-\phi(x)&=\frac{2x}{1+x}\\ 1+(\phi(x))^2&=\frac{2(1+x^2)}{(1+x)^2}\\ \end{align}

Observe that $1+x,1-x,1+x^2,1-x^2,x$ are transformed into product/quotient of these same expressions. It's often nice when dealing with functions containing factors $\ln(1+x),\ln(1+x^2),\ln(x),\ln(1-x)$


this change of variable is linked to $y=\dfrac{2x}{1+x}$


Important thing: $\phi^{-1}=\phi$


Important thing:

If $x\in [0;1]$,

\begin{align}\arctan\left(\phi(x)\right)=\frac{\pi}{4}-\arctan x\end{align}


\begin{align} \phi(\tan x)&=\tan\left(\frac{\pi}{4}-x\right) \end{align}

$y=\phi(x)$ is the equivalent in the "algebraic side" of the trigonometric substitution $y=\dfrac{\pi}{4}-x$ for integrals $\displaystyle \int_0^{\frac{\pi}{4}}f(\tan x)dx$


The so-called Serret integral $\displaystyle \int_0^1 \dfrac{\ln(1+x)}{1+x^2}\,dx$ wasn't first computed (at least in a text published) by Serret (see the article mentioned in Omegadot's answer)

Nowdays people copies and pastes texts from other people but often they don't check out the sources themselves.

If you search for the article of Serret (1844):


The beginning of this article (english translation from French) says (sorry for my bad English):

"The value of this integral, given by M. Bertrand in the preceding volume of this journal, could be obtained quickly without computing any integral."

(preceding volume published in 1843)

The article of M. Bertrand about the so-called Serret integral: http://sites.mathdoc.fr/JMPA/PDF/JMPA_1843_1_8_A7_0.pdf

The method used is not easy to understand for sure. But it gives the good result.

  • $\begingroup$ Thanks for the Addendum. $\endgroup$
    – omegadot
    Mar 2 '19 at 5:44
  • $\begingroup$ A fantastic solution. Thank you very much for the depth you went into in your example. Very much appreciated. $\endgroup$
    – user150203
    Mar 3 '19 at 5:16
  • $\begingroup$ Great answer. In PS3 I believe it should be $\phi(\tan x) = \tan(\pi/4 - x)$. $\endgroup$
    – eccheng
    Mar 3 '19 at 10:27
  • $\begingroup$ Eccheng: You're right. Thanks. $\endgroup$
    – FDP
    Mar 3 '19 at 16:01

I think the substitution you mean is the so-called self-similiar substitution of $$x = \frac{1 - t}{1 + t}.$$ Notice that such a substitution is equal to its own inverse and is therefore an example of an involution.

Such a substitution works best on definite integrals when the limits of integration are between 0 and 1 and the integrand contains factors involving $x$, $1- x$, and $1 + x$ (and including terms that reduce to these through factorisation like $1 - x^2$ and so on), but naturally it is by no means limited to such integrals. Again, as with anything to do with integration, it comes down to practice.

You may find the short expository paper Finding some integrals using an interesting self-similar substitution of interest as it contains a number of interesting examples that make use of the substitution.

As a more difficult example, you may like to cut your teeth on the following: $$\int_0^1 \frac{1}{1 - x^2} \ln \left (\frac{1 + x}{2x} \right ) \, dx = \frac{\pi^2}{12}.$$

  • 2
    $\begingroup$ The change of variable $y=\dfrac{2x}{1+x}$ gives the result quickly. When you have $\ln(f(x))$ first thing to do is to checkout if you can perform the changes of variable $y=f(x)$ or $y=\dfrac{1}{f(x)}$ $\endgroup$
    – FDP
    Mar 1 '19 at 12:15
  • $\begingroup$ Thanks for the article. (I think the historical background is not exact about the so-called Serret integral) $\endgroup$
    – FDP
    Mar 1 '19 at 12:30
  • $\begingroup$ @FDP - Indeed, there are many ways to skin a cat. It seems Paul Nahin, in his book Inside interesting integrals refers to the integral $$\int_0^1 \frac{\ln (1 + x)}{1 + x^2} \, dx,$$ as Serret's integral. Is this the historical inaccuracy you are referring to in the paper? $\endgroup$
    – omegadot
    Mar 1 '19 at 12:50
  • $\begingroup$ Omegadot: too lenghty for a comment see my answer updated $\endgroup$
    – FDP
    Mar 1 '19 at 15:20
  • $\begingroup$ A very warm and massive appreciate for your post and it's detail. I look forward to going through the document you cited. Again, very much appreciated. $\endgroup$
    – user150203
    Mar 3 '19 at 5:16

In my opinion, this substitution is just a small child, or a span of much bigger substitution.

See for example this case, indeed your substitution would work quite nicely if from my link we had $\ln(1+x)$ and $P(x)$ would be something like: $x^2+1$ or $x^2+4x+5$ or other variations, but those are just a few cases.

I don't think that one can answer this rigorously as it's quite subjective, you can tackle the integral by a different method than with this substitution (sometimes it becomes much harder), but indeed it's quite useful when it comes to solve integrals by symmetry, which I believe you should aim to look at when you want to apply this substitution mostly (if you believe that symmetry is a way, then do it) and usually this will happen when you have either an arctan function or logarithm in the numerator combined with a rational function.

A beautiful use of it is done by Jack in this answer, which is quite amazing how everything simplies, but as I mentioned just look for symmetry. I also tend to use it when I have an odd degree polynomial in the denominator, because it simplifies quite nice to a lower degree polynomial, see for example what I am talking about here and here.

  • 1
    $\begingroup$ Excellent solution. I am very fortunate to pose a question that receives brilliant answers all around. I wish I could give you all the 'answer'. $\endgroup$
    – user150203
    Mar 3 '19 at 5:17

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