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Evaluate: $\displaystyle\int_{0}^{1}\frac{\ln(x+1)}{x^2+1}\,\mathrm{d}x$

So I did this a completely different way than what the answer key states. I used integration by parts and some symmetry tricks and got the correct answer. However the answer key says:

Make the substitution $x=\frac{1-u}{1+u}$

The same solution was reached in about half the steps but still using symmetry, my questions are How would I know to do that? Is this a certain type of substitution? Is there something else that maybe I could use this for?

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  • $\begingroup$ the change of variable reminds me a bit the Weierstrass substitution $\cos(x)=\frac{1-t^2}{1+t^2}$ $\endgroup$ – Masacroso May 18 '17 at 20:31
  • $\begingroup$ That is true. The weierstrauss sub is actually what I used in part of my long solution $\endgroup$ – Teh Rod May 18 '17 at 20:33
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    $\begingroup$ The transformation is a Möbius Transform, not a Weierstrass substitution. $\endgroup$ – Mark Viola May 18 '17 at 20:39
  • $\begingroup$ possible duplicate math.stackexchange.com/questions/155941/… $\endgroup$ – Darío A. Gutiérrez May 18 '17 at 21:09
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The substitution is called a Möbius Transformation, which has the more general form

$$w=\frac{az+b}{cz+d}$$

where $ad\ne bc$. The transformation maps straight lines into circles and circles into straight lines. They have a variety of uses in applied mathematics and physics along with complex analysis.


As a simple example, the Beta function $B(x,y)$ can be represented by the integral

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$

Enforcing the Mobius Transformation $t\to \frac{t}{1+t}$ so that $dt\to \frac{1}{(1+t)^2}\,dt$. Then, we see that

$$B(x,y)=\int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}\,dt$$

which is an alternative representation for the Beta function.


As another example, in This Question, the OP requested evaluation of the integral $I$ expressed as

$$I=\int_0^\infty\frac{\log(e^x-1)}{e^x+1}\,dx$$

In the accepted answer posted by User @FDP, the Mobius transform was used to facilitate an efficient way forward.


As a third and final example, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x}$$

for $x<1$. Then, it is trivial to see that $\log(x)\le x-1$.

Applying the Möbius Tranformation $x\to \frac{-x}{1+x}$ we find that the logarithm function is bounded below by $\log(x)\ge \frac{x-1}{x}$.

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Here's an alternative method that doesn't use differentiation under the integral sign. Substitute $x = \tan \theta$, to give $$I = \int_0^{\pi/4} \log(1+\tan\theta) d\theta \\ = \int_0^{\pi/4} \log(\cos\theta+\sin\theta)-\log\cos(\theta)d\theta \\ = \int_0^{\pi/4} \frac12\log2 + \log(\cos(\theta- \pi/4))-\log\cos(\theta) d\theta \\ = \frac{\pi}{8}\log2 + \int_0^{\pi/4} \log(\cos(\theta- \pi/4)) d\theta - \int_0^{\pi/4}\log\cos(\theta) d\theta $$ We have the known property that $\int_0^a f(x) dx = \int_0^a f(a-x)dx$, and $\cos$ is an even function, so these two final integrals are equal, and $I = \frac{\pi}{8}\log2$.

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Here's how I would do it. Let $$f(\alpha)=\int_0^1 \frac{\log(1+\alpha x)}{1+x^2}\ dx$$ where $f(1)$ is the integral we seek to evaluate. By differentiation under the integral sign we have $$f'(\alpha)=\int_0^1\frac{\partial}{\partial\alpha}\frac{\log(1+\alpha x)}{1+x^2}\ dx\\ =\int_0^1\frac{x}{(1+x^2)(1+\alpha x)}\ dx.$$

Now we apply partial fraction decomposition: suppose that $$\frac{x}{(1+x^2)(1+\alpha x)}=\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}$$ for some constants $A,B,C$. By combining the fractions and collecting the terms we obtain the linear system of equations $$\begin{cases}A+\alpha B=1\\ \alpha A+C=0\\B+C=0\end{cases}$$ which we find to be equivalent to $$\begin{cases}A=\frac{1}{1+\alpha^2}\\B=\frac{\alpha}{1+\alpha^2}\\C=-\frac{\alpha}{1+\alpha^2}\end{cases}$$ From this we conclude that $$f'(\alpha)=\int_0^1\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}\ dx\\ =\left[\frac12 A\log(1+x^2)+B\arctan{x}+\frac{C}{\alpha}\log(1+\alpha x)\right]_{x=0}^1\\ =\frac{\log2}{2}\frac{1}{1+\alpha^2}+\frac{\pi}{4} \frac{\alpha}{1+\alpha^2}-\color{red}{\frac{\log(1+\alpha)}{1+\alpha^2}}$$ It is easy to see by looking at the definition that $f(0)=0$. From this it follows that $f(\alpha_0)=\int_0^{\alpha_0} f'(\alpha)\ d\alpha$, and in particular $$f(1)=\int_0^1 \frac{\log2}{\alpha}\frac{1}{1+\alpha^2}+\frac{\pi}{4}\frac{\alpha}{1+\alpha^2}-\color{red}{\frac{\log(1+\alpha)}{1+\alpha^2}}\ d\alpha$$ which simplifies to $$\color{red}{\int_0^1 \frac{\log(1+x)}{1+x^2}\ dx}= \frac{\pi \log 2}{16}-\color{red}{\int_0^1 \frac{\log(1+\alpha)}{1+\alpha^2}\ d\alpha}$$ hence the result $$\int_0^1 \frac{\log(1+x)}{1+x^2}\ dx=\frac{\pi \log 2}{8}.$$

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  • $\begingroup$ That is a smarter way of doing it are you sure you meant $\frac{\log 2}{x}$? $\endgroup$ – Teh Rod May 18 '17 at 20:45
  • $\begingroup$ @TehRod that was a typo, now amended :) $\endgroup$ – user1892304 May 18 '17 at 20:51
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Set $$x=\tan t, \qquad dt=\dfrac1{1+x^2}dx,$$ Then $$ \begin{align} \int_0^1 \frac{\ln(1+x)}{1+x^2}\, dx&= \int_0^{\pi/4} \ln(\cos t +\sin t)\, dt- \int_0^{\pi/4} \ln(\cos t)\, dt\\\\ &=\int_0^{\pi/4} \ln\left(\sqrt{2}\cos \left(\frac \pi4- t\right)\right)\, dt- \int_0^{\pi/4} \ln(\cos t)\, dt\\\\ &=\int_0^{\pi/4} \ln\left(\sqrt{2}\right)\, dt+\int_0^{\pi/4} \ln\left(\cos \left(\frac \pi4- t\right)\right)\, dt- \int_0^{\pi/4} \ln(\cos t)\, dt\\\\ &=\frac{\pi}8 \:\ln 2+\int_0^{\pi/4} \ln(\cos u)\, du-\int_0^{\pi/4} \ln(\cos t)\, dt\quad \left(u=\frac \pi4- t\right)\\\\ &=\frac{\pi}8 \:\ln 2. \end{align} $$

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