# Why aren't these two integrals equivalent when using the substitution $x=\frac{1}{t}$?

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 $$1$$?

• should be $dx=-\dfrac{1}{t^2}dt$ – janmarqz Mar 7 at 16:25
• 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 $(1,\infty)$ – Ninad Munshi Mar 7 at 16:25
• @janmarqz OP did the substitution correctly (except for the bounds issue I mentioned) – Ninad Munshi Mar 7 at 16:28

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 $$\infty$$.)

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

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

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 $$-\infty.$$