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$$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$

I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.

$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)$$
So $$I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty $$ which evaluates to be $\infty$.

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    $\begingroup$ Try 1/x - x/(x^2+1) $\endgroup$ – jnyan Nov 23 '17 at 6:03
  • $\begingroup$ The factor $\frac 12$ does not apply for $\frac 1x$. Otherwise, you are on the right track if you simplify the complex terms. $\endgroup$ – Claude Leibovici Nov 23 '17 at 6:07
  • $\begingroup$ @ClaudeLeibovici How about if I don't simplify the complex terms (just substituting the limits in)? Wouldn't I get something like $-\infty - \infty$ (a sum of infinities)? $\endgroup$ – Natash1 Nov 23 '17 at 6:11
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    $\begingroup$ You end up with $$ I = \int_{1}^{\infty} \left( \frac{1}{x} - \frac{x}{1+x^2} \right) \, dx = \int_{1}^{\infty} \left( \frac{1}{x} - \frac{1}{2(x+i)} - \frac{1}{2(x-i)} \right) \, dx. $$ And notice that this is the limit $$ I = \lim_{R\to\infty} \int_{1}^{R} \left( \frac{1}{x} - \frac{1}{2(x+i)} - \frac{1}{2(x-i)} \right) \, dx. $$ This does help in evaluating $I$ since you now have a direct control over the indeterminate of the form $\infty - \infty$. $\endgroup$ – Sangchul Lee Nov 23 '17 at 6:15
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    $\begingroup$ Possible duplicate of Evaluation of $\int_1^\infty \frac{1}{x(x^2+1)} dx$ $\endgroup$ – Nosrati Dec 11 '18 at 11:06
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Why complicate it with complex coefficients? $$\int_1^\infty\frac{1}{x(x^2+1)}dx=\int_1^\infty\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx=\left(\ln(x)-\frac{1}{2}\ln(x^2+1)\right)\bigg\vert_1^\infty\\=\ln\frac{x}{\sqrt{x^2+1}}\bigg\vert_1^\infty=\ln\frac{1}{\sqrt{1+1/x^2}}\bigg\vert_1^\infty=\ln 1-\ln\frac{1}{\sqrt{2}}=\ln\sqrt 2$$

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  • $\begingroup$ While your answer is correct, I think it'd be more helpful to show OP what was wrong in their method $\endgroup$ – Dylan Nov 23 '17 at 7:33
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First of all: $$\frac{1}{x(x^2+1)} = \frac{1}{x}-\frac{1}{2}\left(\frac{1}{x+i} -\frac{1}{x-i}\right)$$

Second of all:

You have to evaluate the integral $\int_1 ^ R\frac{1}{x(x^2+1)}dx$ then take the limit as R goes to infinity and note that $\infty -\infty$ is undetermined.

$$\int_1 ^ R\frac{1}{x(x^2+1)} = \int_1 ^ R \left[\frac{1}{x}-\frac{1}{2}\left(\frac{1}{x+i} -\frac{1}{x-i}\right)\right]dx$$ $$=[\log |x| - \frac{1}2{}\log(|x+i|) - \frac{1}{2}\log(|x-i|)]_1^R$$

$$=[\log |R| - \frac{1}2{}\log(|R+i|) - \frac{1}{2}\log(|R-i|)] + \frac{1}2{}\log(|1+i|) + \frac{1}{2}\log(|1-i|)]$$

$$=\log\left|\frac{R}{(R+i)^{\frac{1}{2}}(R-i)^\frac{1}{2}}\right|+\frac{1}{2}\log(1+i)(1-i)$$ $$=\log\left|\frac{R}{(R^2+1)^\frac{1}{2}}\right|+\frac{1}{2}\log(2)$$ Now taking the limit yields $\frac{1}{2}\log(2)$.

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put $x = \tan \theta$ given integral is $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sec^2 \theta}{\tan \theta \sec^2 \theta} = \frac{d(\sin \theta)}{\sin \theta} = \ln [|\sin \theta|]^{\frac{\pi}{2}}_{\frac{\pi}{4}} = 0-\ln \frac{1}{\sqrt{2}}=\ln \sqrt{2}$$

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Your partial fraction decomposition is incorrect. If you still want to use complex numbers, it's

$$ \frac{1}{x(1+x^2)} = \frac{1}{x} - \frac{1}{2(x+i)} - \frac{1}{2(x-i)} $$

Then

$$ \int_1^{\infty} f(x) \ dx = \left.\left(\ln |x| - \frac{1}{2}\ln |x+i| - \frac{1}{2} \ln |x-i|\right)\right|_1^{\infty} = \left. \ln \left|\frac{x}{\sqrt{x^2+1}} \right|\right|_1^{\infty} = \frac{1}{2}\ln 2 $$

Note that $$ \lim_{x\to \infty} \frac{x}{\sqrt{x^2+1}} = \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}} = 1 $$

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    $\begingroup$ The limit at $1$ is not $0$. How can the integral of a positive function be $0$? $\endgroup$ – Andrei Nov 23 '17 at 8:09
  • $\begingroup$ Thanks. For some reason I thought it evaluates to $\ln 1$. I've updated my answer $\endgroup$ – Dylan Nov 23 '17 at 9:08
  • $\begingroup$ No problem. Happens to everyone from time to time $\endgroup$ – Andrei Nov 23 '17 at 9:20
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By enforcing the substitution $x=\frac{1}{z}$ the result is straightforward to find:

$$ \int_{1}^{+\infty}\frac{dx}{x(x^2+1)}=\int_{0}^{1}\frac{z\,dz}{z^2+1}=\left[\frac{1}{2}\log(1+z^2)\right]_{0}^{1}=\color{red}{\frac{\log 2}{2}}.$$

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