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How can I evaluate this improper integral?

$$\displaystyle\int_0^{\infty}\frac{1}{x(1+x^2)}\,dx $$

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This doesn't converge. –  Pedro Tamaroff May 9 '13 at 2:35
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Near $x=0$ your function is of the same order as $1/x$, so the left limit of integration is a problem for convergence. –  alex.jordan May 9 '13 at 2:37
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4 Answers

$$\int_0^1\frac{1}{x(1+x^2)}\,dx\geq\int_0^1\frac1{2x}\,dx$$ which is more clearly divergent ($\lim\limits_{t\to0^+}\left[\frac{1}{2}\ln(x)\right]_t^1$).

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Try this:

$$x=\tan { u } $$ $$dx=\frac { 1 }{ { \left( \cos { u } \right) }^{ 2 } } du$$ $$\int _{ 0 }^{ \infty }{ \frac { 1 }{ x\left( { x }^{ 2 }+1 \right) } dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos { u } }{ \sin { u } } du } $$

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Thank you, I tried that and it's really easy to see that the integral is divergent. –  Lia Denisse May 9 '13 at 3:08
    
you're wellcome. –  newzad May 9 '13 at 3:14
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Help getting started: Start by writing the integral like this: $$ \int_0^\infty \frac{1}{x(1+x^2)}\; dx = \lim_{t\to 0^+}\int_t^1\frac{1}{x(1+x^2)}\; dx + \lim_{s\to \infty}\int_1^{s}\frac{1}{x(1+x^2)}\; dx. $$ If you can show that one of these integrals is divergent, then the original integral is divergent.

First we can consider the integral $$\begin{align} \int_t^1 \frac{1}{x(1+x^2)}\; dx &= \int_t^1 \frac{1}{x} - \frac{x}{1+x^2}\; dx \\ &= \ln\lvert x\rvert - \frac{1}{2}\ln\lvert 1 + x^2\rvert\left. \right]_t^1\\ &= \ln\left(\frac{x}{\sqrt{1+x^2}}\right)\left.\right]_t^1 \end{align} $$ I believe that when you evaluate this and take the limit as $t$ goes to $0$ from the right, then is $-\infty$. So yes, the integral is divergent.

But, you should probably check the details.

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Ok, so if I use $ \lim_{t\to 0^+}\int_t^1\frac{1}{x(1+x^2)}\; dx = \log 1 -\lim_{t\to 0^+}\log x- \frac{1}{2}\log 2 + \frac{1}{2}\log 1 = \infty $ then the original integral is divergent? –  Lia Denisse May 9 '13 at 2:56
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An indefinite integral leads to (breaking it into simple fractions) $$log(x)-\frac{1}{2}log(1+x^2)$$ So clearly you have problems at $x=0$.

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