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Why aren't these two integrals $$\int_{-1}^{1}\frac{1}{\left(1+x^2\right)^2}\,\mathrm{d}x$$ and $$\int_{-1}^{1}\frac{-t^2}{\left(1+t^2\right)^2}\,\mathrm{d}t$$ equal to each other, despite using the substitution $x=\frac{1}{t}$, which yields the second integral when the substitution is used on the first one?

Could it be that $t$ is undefined at $x=0$ since the limits are from $-1$ to $1$?

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  • $\begingroup$ should be $dx=-\dfrac{1}{t^2}dt$ $\endgroup$ – janmarqz Mar 7 at 16:25
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    $\begingroup$ You had to break integral up into two pieces to see why. The substitution is not defined at $0$. The correct substitution gets an integral over $(-\infty,-1)$ and $(1,\infty)$ $\endgroup$ – Ninad Munshi Mar 7 at 16:25
  • $\begingroup$ @janmarqz OP did the substitution correctly (except for the bounds issue I mentioned) $\endgroup$ – Ninad Munshi Mar 7 at 16:28
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The substitution is not defined for $x=0$ or at $t=0$. So in fact if you make the substitution, what you have is

$$\begin{align} \int_{-1}^1{1\over(1+x^2)^2}dx&=\int_{-1}^0{1\over(1+x^2)^2}dx+\int_0^1{1\over(1+x^2)^2}dx\\ &=\int_{-1}^{-\infty}{-t^2\over(1+t^2)^2}dt+\int_\infty^1{-t^2\over(1+t^2)^2}dt\\ &=\int_{-\infty}^{-1}{t^2\over(1+t^2)^2}dt+\int_1^{\infty}{t^2\over(1+t^2)^2}dt \end{align}$$

(And as J.G. points out, the symmetry of the integrand allows you to reduce to just one integral, from $1$ to $\infty$.)

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If $x=\frac1t$, then, since $x\in[-1,1]$, you should get$$\int_{-\infty}^{-1}\frac{t^2}{(1+t^2)^2}\,\mathrm dt+\int_1^\infty\frac{t^2}{(1+t^2)^2}\,\mathrm dt.$$

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For even $f$,$$\int_{-1}^1f(x)\mathrm{d}x=2\int_0^1f(x)\mathrm{d}x=2\int_1^\infty\tfrac{f(1/t)}{t^2}\mathrm{d}t.$$

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As $x$ goes from $0$ to $+1,$ $t$ goes from $+\infty$ to $1.$

As $x$ goes from $-1$ to $0,$ $t$ goes from $-1$ to $-\infty.$

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