3

The integral is:

$$\int_{0}^{1} \dfrac{\ln(x+1)}{x^2+1} dx$$

This is a Putnam question from my calculus book. It was in the section about trig substitution. However, I see no way of how mixing a trig function with a natural logarithm would make it any easier. I've tried the substitution $x=\tan \theta$ and got $$\dfrac 12\int \ln(1+\tan \theta)^2 d \theta$$

However, I wasn't able to simplify it to any integrable form.

Ovi
  • 23,737
  • 1
    The fact that the original integral is definite suggests that the integrand might not have a closed-form antiderivative (as I think is the case). So, one should probably carry through the bounds and use the symmetry of the integrand to evaluate the integral. – Travis Willse Apr 20 '16 at 21:21
  • @learner That's just the substitute $x = \tan \theta$ that OP used to rewrite the integral. – Travis Willse Apr 20 '16 at 21:21

4 Answers4

8

One may write $$ \begin{align} \int_0^1\frac{\ln (1+x)}{1+x^2}\:dx&=\int_0^{\pi/4} \ln (1+\tan u)\:du \quad \left(x=\tan u, \,du=\frac{dx}{1+x^2}\right) \\\\&=\int_0^{\pi/4} \ln \left(1+\tan \left(\frac\pi4-v\right)\right)\:dv\qquad (u=\frac\pi4-v ) \\\\&=\int_0^{\pi/4} \ln \left(1+\frac{1-\tan v}{1+\tan v}\right)dv \quad \left(\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b}\right) \\\\&=\int_0^{\pi/4} \ln \left(\frac2{1+\tan v}\right)dv \\\\&=\int_0^{\pi/4} \ln 2\:dv-\int_0^{\pi/4}\!\! \ln (1+\tan v)\:dv \end{align} $$ giving

$$ \int_0^1\frac{\ln (1+x)}{1+x^2}\:dx=\int_0^{\pi/4}\!\! \ln (1+\tan u)\:du=\frac\pi8 \: \ln 2. $$

Olivier Oloa
  • 120,989
  • I have seen this techinque used before. Does it have a name? Is it a standard technique learned in a class? – Ovi Apr 20 '16 at 22:40
  • I don't know if it has a name, but the idea is to exploit symmetry as much as possible. Thanks. – Olivier Oloa Apr 21 '16 at 02:18
  • Are you talking about symmetry in the equations? Because graphically, neither $\dfrac{\ln(1+x)}{1+x^2}$ nor $\ln(1+\tan u)$ look symmetrical on the given interval. – Ovi Apr 21 '16 at 02:56
  • Also, was there a specific reason you were motivated you to make the substitution $u=\pi/4-v$ and use the difference formula? I have seen this technique before and I want to pick it up. But besides a few isolated examples on stackexchange I haven't been able to find sources that give the technique a more general treatment. – Ovi Apr 21 '16 at 02:57
  • The following symmetry: $\ln (1+ \tan ) \circ (\pi/4-x) =\ln 2- \ln (1+ \tan x)$ helped us. I've played around with this technique when I was younger. I don't know if it is collected somewhere. – Olivier Oloa Apr 21 '16 at 03:55
  • Hm, so from what I've gathered so far about the technique, the function doesn't really have to be symmetric at all. If we have $\int_{a}^{b} f(x) dx$, it seems like all we have to do is make a substitution of $x=u-b-a$ and, and we will have $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(u-b-a)du$. And if we are lucky, we will be able to rewrite the RHS as the difference of some simple integral and $\int_{a}^{b} f(u)du$. I will have to experiment to see if I can make a similar argument with the original. It might be impossible, but I don't see any reason to immediately rule it out as such – Ovi Apr 21 '16 at 05:21
5

It is enough to exploit the substitution $x=\tan t$ and the identity $\sin t+\cos t=\sqrt{2}\cos\left(\frac{\pi}{4}-t\right)$:

$$\begin{eqnarray*}I=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\,dx &=& \int_{0}^{\pi/4}\log(\sin t+\cos t)-\log(\cos t)\,dt\\&=&\frac{\pi}{8}\log(2)+\int_{0}^{\pi/4}\log\cos\left(\frac{\pi}{4}-t\right)-\log\cos t\,dt\\&=&\frac{\pi}{8}\log(2)+\int_{0}^{\pi/4}\log\cos t\,dt-\int_{0}^{\pi/4}\log\cos t\,dt\\&=&\color{red}{\frac{\pi}{8}\,\log 2}.\end{eqnarray*}$$

Jack D'Aurizio
  • 353,855
4

A substitution other than a trigonometric substitution can be used to evaluate this integral.

Using a so-called self-similar substitution of $u = \dfrac{1 - x}{1 + x}$, one has $x = \dfrac{1 - u}{1 + u}$. Thus $dx = -\dfrac{2}{(1 + u)^2} \, du$ and for the limits of integration when $x = 0, u = 1$ and when $x = 1, u = 0$.

Noting that $1 + x = \dfrac{2}{1 + u}$ and $1 + x^2 = \dfrac{2(1 + u^2)}{(1 + u)^2}$ the integral becomes \begin{align*} I &= \int^1_0 \frac{\ln (1 + x)}{1+ x^2} \, dx\\ &= \int^1_0 \ln \left (\frac{2}{1 + u} \right ) \cdot \frac{(1 + u)^2}{2 (1 + u^2)} \cdot \frac{2}{(1 + u)^2} \, du\\ &= \int^1_0 \ln \left (\frac{2}{1 + u} \right ) \frac{du}{1 + u^2}\\ &= \int^1_0 \left [\ln 2 - \ln (1 + u) \right ] \frac{du}{1 + u^2}\\ &= \ln 2 \int^1_0 \frac{du}{1 + u^2} - \int^1_0 \frac{\ln (1 + u)}{1 + u^2} \, du\\ \Rightarrow 2I &= \ln 2 \cdot \Big{[}\tan^{-1} u \Big{]}^1_0 \end{align*}
giving $$\int^1_0 \frac{\ln (1 + x)}{1 + x^2} \, dx = \frac{\pi}{8} \ln 2.$$

omegadot
  • 11,736
3

Instead of diverse substitutions we can study the function $$ f(s) = \int_{0}^{1}\dfrac{\ln(1+sx)}{1+x^2}\, dx $$ where we are interested in $f(1)$. At first we determine \begin{gather*} f'(s) = \int_{0}^{1}\dfrac{x}{(1+sx)(1+x^2)}\, dx = \int_{0}^{1}\dfrac{1}{1+s^{2}}\left(\dfrac{-s}{1+sx}+\dfrac{x+s}{1+x^{2}}\right)\, dx \\[2ex] = -\dfrac{\ln(1+s)}{1+s^{2}} + \dfrac{\ln(2)}{2(1+s^{2})} + \dfrac{s\pi}{4(1+s^{2})}. \end{gather*} Since $f(0)=0$ $$ f(1) = \int_{0}^{1}f'(s)\, ds = -\int_{0}^{1}\dfrac{\ln(1+s)}{1+s^{2}}\, ds + \dfrac{\pi\ln 2}{4} = -f(1) + \dfrac{\pi\ln 2}{4}. $$ Consequently $$ f(1) = \dfrac{\pi\ln 2}{8}. $$

JanG
  • 4,577