We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
165
2
avatar

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

 Jun 5, 2018
 #1
avatar+974 
+2

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
avatar+21978 
+2

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}\)

 

 

laugh

 Jun 5, 2018

3 Online Users