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Putnam Competition Clarification

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$$\int_0^{1}\frac{\ln(x+1)}{x^2+1}dx$$

Jun 5, 2018

#1
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Hey Guest,

I have slept on it, and finished the solution.

First, we use $$x = \tan θ$$

We can rewrite the desired integral as $$\int_0^{\pi/4}\log(\tan(\theta)+1)d\theta$$

Also:

$$\log(\tan(θ) + 1) = \log(\sin(θ) + \cos(θ)) − \log(\cos(θ))$$

Note that: $$\sin(θ) + \cos(θ) = \sqrt2 \cos(π/4 − θ).$$

Rewriting the integral, we get:

$$\frac12\log(2) + \log(\cos(π/4 − θ)) − \log(\cos(θ)).$$

Over the interval $$[0, π/4]$$, the integrals of $$\log(\cos(θ))\ \text{and}\ \log(\cos(π/4 − θ))$$ are equal, so their contributions cancel out. The answer you seek is $$π \log(2)/8.$$

Gavin

Edit: I have done some research, and googled Putnam competition.

This problem is A5 of the 66th annual competition.

The full solution is here: http://math.hawaii.edu/home/pdf/putnam/2005.pdf.

Jun 5, 2018
edited by GYanggg  Jun 5, 2018
edited by GYanggg  Jun 5, 2018
#2
+22884
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Integral

\displaystyle \int \limits_0^{1}\frac{\ln(x+1)}{x^2+1}dx

$$\displaystyle \int \limits_0^{1}\frac{\ln(x+1)}{x^2+1}dx$$

$$\begin{array}{|llcl|} \hline \displaystyle \int \limits_{x=0}^{x=1}\dfrac{\ln(x+1)}{x^2+1}\ dx =\ ? \\\\ \text{substitute:}\\ & x = \tan(z) \\ & dx =( \tan^2(z)+1)\ dz \\ \text{new limits:}\\ & x=0 \Rightarrow z = \arctan(0) = 0 \\ & x=1 \Rightarrow z = \arctan(1) = \frac{\pi}{4} \\ \hline \end{array}$$

$$\begin{array}{|rcl|} \hline && \displaystyle \int \limits_{x=0}^{x=1}\dfrac{\ln(x+1)}{x^2+1}\ dx \\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\dfrac{\ln(\tan(z)+1)}{\tan^2(z)+1} ( \tan^2(z)+1)\ dz \\\\ &=& \color{red}\displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(\tan(z)+1) \ dz \\\\ && \boxed{\text{Formula: }\int_0^a f(x)dx=\int_0^a f(a-x)dx} \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln\left(\tan(\frac{\pi}{4}-z)+1 \right) \ dz \\\\ && \boxed{\text{Formula: } \\ \tan(\frac{\pi}{4}-z) = \frac{\tan(\frac{\pi}{4})-\tan(z) }{1+\tan(\frac{\pi}{4})\tan(z)} \\ = \frac{1-\tan(z)}{1+\tan(z)} } \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln\left(\frac{1-\tan(z)}{1+\tan(z)}+1 \right) \ dz \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln\left(\frac{1-\tan(z)+1+\tan(z)}{1+\tan(z)} \right) \ dz \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln\left(\frac{2}{1+\tan(z)} \right) \ dz \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(2)\ dz - \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(1+\tan(z) ) \ dz \\\\ &=& \displaystyle [\ln(2)z]_{z=0}^{z=\frac{\pi}{4}} - \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(1+\tan(z) ) \ dz \\\\ &=& \displaystyle \ln(2)\frac{\pi}{4} - \color{red}\int \limits_{z=0}^{z=\frac{\pi}{4}}\ln (1+\tan(z) ) \ dz \\\\ \color{red}\displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(\tan(z)+1) \ dz &=& \displaystyle \ln(2)\frac{\pi}{4} - \color{red}\int \limits_{z=0}^{z=\frac{\pi}{4}}\ln (1+\tan(z) ) \ dz \\\\ 2\cdot \color{red}\displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(\tan(z)+1) \ dz &=& \displaystyle \ln(2)\frac{\pi}{4} \\\\ \color{red}\displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(\tan(z)+1) \ dz &=& \displaystyle \ln(2)\frac{\pi}{8} \\\\ \displaystyle \int \limits_{x=0}^{x=1}\dfrac{\ln(x+1)}{x^2+1}\ dx &=& \displaystyle \ln(2)\frac{\pi}{8} \\\\ \mathbf{\displaystyle \int \limits_{x=0}^{x=1}\dfrac{\ln(x+1)}{x^2+1}\ dx }& \mathbf{=} & \mathbf{0.27219826129} \\ \hline \end{array}$$

Jun 5, 2018