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While solving the integral $$\int_2^\infty\frac{2}{t^2-1} \; dt,$$ I encountered one of the limits according to multiple sites is $i\pi$ and $\ln(3)$. But in the final answer, only $\ln(3)$ is listed. I am having trouble understanding how the limit $i\pi$ is found. Thank you in advance

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5  
You're integrating a real function on the real line; you obviously can't get an imaginary answer. – T. Bongers Mar 1 at 5:55
    
Thank you for clarifying, when finding the definite integral, the imaginary number can be disregarded? – Marc D. Mar 1 at 6:16
2  
Not really, I'm saying that it makes no sense in the first place. I have no idea where one would get $i\pi$ out of this. – T. Bongers Mar 1 at 6:17
    
I took the limit of -ln((1+x)/(1-x)) as it approached infinity and two. By doing this it produced an imaginary limit. – Marc D. Mar 1 at 6:28
up vote 3 down vote accepted

We have $$\int_{2}^{\infty}\frac{2}{t^{2}-1}dt = -2\int_{2}^{\infty}\frac{1}{1-t^{2}}dt = -2\left[\tanh^{-1}\left(t\right)\right]_{2}^{\infty} $$ and $$\lim_{t\rightarrow\infty}\tanh^{-1}\left(t\right)=-\frac{1}{2}i\pi $$ and $$-2\tanh^{-1}\left(2\right) = -\left(\log\left(3\right)-\log\left(-1\right)\right) = -\log\left(3\right)+i\pi $$ so $$\int_{2}^{\infty}\frac{2}{t^{2}-1}dt = \log\left(3\right). $$

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So we disregard $i\pi$ as part of the solution? – Marc D. Mar 1 at 6:31
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@MarcD. The part $i\pi$ is deleted from the calculations. – Marco Cantarini Mar 1 at 6:34
    
Thank you for your answer! – Marc D. Mar 1 at 6:35
5  
@MarcD., isn't disregarded. $i\pi - (-\log(3) +i\pi) = \log(3)$. – Martín-Blas Pérez Pinilla Mar 1 at 9:37

Why not to simply use partial fraction decomposition and write$$\int\frac{2}{t^2-1} \; dt=\int\Big(\frac{1}{t-1}-\frac{1}{t+1}\Big)\,dt=\log\Big(\frac{t-1}{t+1}\Big)$$ which makes $$\int_2^\infty\frac{2}{t^2-1} \; dt=\log(1)-\log\left(\frac 13\right)=\log(3)$$

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