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I have a function $y(x)=\frac{x}{1+x^2}, x\in Reals$, it is an odd function, with no undefined points, so I expect its integral (in the range $[-\infty,+\infty]$) to be 0. But when I use Mathematica to calculate it,

Integrate[x/(1+x^2),{x,-Infinity,Infinity}]

It says 'Integral of $y(x)=\frac{x}{1+x^2}$ does not converge on $[-\infty,+\infty]$'.

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  • $\begingroup$ Also, Integrate[x/(1+x^2), {x, -max, max}] evaluates to zero and its limit as max -> Infinity is also zero. $\endgroup$ – Bob Hanlon Dec 27 '18 at 14:59
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Integrate[x/(1 + x^2), {x, -Infinity, Infinity}, 
 PrincipalValue -> True]

Out[1]= 0
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  • $\begingroup$ Hi, thanks, I tried this and it works. But could you explain a little bit more about the 'principalvalue'? $\endgroup$ – dr.bian Dec 27 '18 at 14:33
  • $\begingroup$ @dr.bian It is well known, if the Riemann integral diverges, then the Cauchy integral may converge. Your intuition tells you "it is an odd function, with no undefined points, so I expect its integral to be 0". Mathematica answers you "the Cauchy principal value integral =0". $\endgroup$ – Alex Trounev Dec 27 '18 at 14:51
  • $\begingroup$ Ah,I will see more on this, Thanks! $\endgroup$ – dr.bian Dec 27 '18 at 15:00
  • $\begingroup$ @dr.bian You're welcome! $\endgroup$ – Alex Trounev Dec 27 '18 at 15:02

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