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Graph of $\frac{x}{x^2 + 1}

I am trying to find the area of this graph $\int_{-\infty}^\infty\frac{x}{x^2 + 1}$

The question first asks to use the u-substitution method to calculate the integral incorrectly by evaluating $\lim \limits_{a \to \infty}\int_{-a}^a\frac{x}{x^2+1}dx$

My approach to this (not sure) is as follows:

$$u=x^2+1$$ $${du}=2x*dx $$ $$\frac12\int_{-\infty}^{\infty}{\frac1u}du$$ $$=\left[\frac12\log(x^2+1)\right]_{-\infty}^\infty$$

But how do I solve that part?

If I am asked to find

the $\int_1^a\frac{x}{x^2+1}dx$ where a is infinity, how do I do that?

Also what is the proper way to evaluate the integral from $-\infty to \infty$

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    $\begingroup$ The area does not exist, or if you prefer is infinite. We need to examine the area from $0$ to $\infty$, and from $-\infty$ to $0$, separately. $\endgroup$ – André Nicolas Jul 29 '14 at 22:38
  • $\begingroup$ When you are asked to do it incorrectly you should get zero. $\endgroup$ – Brad Jul 30 '14 at 7:23
  • $\begingroup$ You don't have to use MathJax to get italics. See editing help. $\endgroup$ – Martin Sleziak Jul 30 '14 at 8:26
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The function $x\mapsto \frac x{1+x^2}$ is odd so the integral $$\int_{-a}^a\frac{x}{1+x^2}dx=0$$ so the given limit is $0$. Notice that even so the first given integral doesn't exist since

$$\frac x{1+x^2}\sim_\infty \frac1x$$ and the integral $$\int_1^\infty \frac{dx}x$$ is undefined.

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  • $\begingroup$ I know it's weird the question is literally asking to "calculate the integral incorrectly" $\endgroup$ – Potato Monster Jul 29 '14 at 22:41
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    $\begingroup$ You don't need to ask how to do it incorrectly (that wouldn't be maths & would be a waste of time) :) $\endgroup$ – Jam Jul 29 '14 at 22:42
  • $\begingroup$ Quick questino added $\endgroup$ – Potato Monster Jul 29 '14 at 22:45
  • $\begingroup$ i require help :p $\endgroup$ – Potato Monster Jul 29 '14 at 22:57
  • $\begingroup$ The last integral is enough for a nice deduction. :+) $\endgroup$ – mrs Jul 30 '14 at 0:11
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One may use the Limit comparison test for seeing that:

$\lim_{x\to+\infty}x^{1}\times f(x)=1$ so $\int_1^{\infty}f(x)dx$ diverges.

This is an alternative way to what Sami indicated in his post.

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When you are asked to compute $\displaystyle\lim_{a \to \infty}\int_{-a}^{a}\dfrac{x}{x^2+1}\,dx$ and $\displaystyle\lim_{a \to \infty}\int_{1}^{a}\dfrac{x}{x^2+1}\,dx$, you should first compute the integral (in terms of $a$) and then take the limit as $a \to \infty$, not the other way around.

Using the substitution $u = x^2+1$, $du = 2x\,dx$, you get $\displaystyle\int\dfrac{x}{x^2+1}\,dx = \dfrac{1}{2}\ln(x^2+1)+C$.

Thus, $\displaystyle\int_{-a}^{a}\dfrac{x}{x^2+1}\,dx = \left[\dfrac{1}{2}\ln(x^2+1)\right]_{-a}^{a} = 0$ and $\displaystyle\int_{1}^{a}\dfrac{x}{x^2+1}\,dx = \dfrac{1}{2}\ln(a^2+1)-\dfrac{1}{2}\ln 2$.

Now, take the limit of these expressions as $a \to \infty$.

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This a typical case of improper definite integral which cannot be evaluted on the common sense, but which is well defined on the sense of Cauchy Principal Value : http://mathworld.wolfram.com/CauchyPrincipalValue.html $$PV\int_{-\infty}^\infty\frac{x}{x^2 + 1} = 0$$

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