Is $\int_{-\infty}^\infty \sum_{g(x)=a}\frac{f(a)}{|g’(x)|} da=\left(\frac1{|g’(a_1)|}+\dots+\frac1{|g’(a_j)|}\right)\int_{-\infty}^\infty f(a)da$?

Question

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A strange integral: $\int_{-\infty}^{+\infty} {dx \over 1 + \left(x + \tan x\right)^2} = \pi.$

@robjohn posts that:

$$\int_{—\infty}^\infty f(g(x))dx=\int_{-\infty}^\infty \sum_{g(x)=a}\frac{f(a)}{|g’(x)|} da$$

Denote $g(a_n)=a$ with $j$ roots:

$$\int_{-\infty}^\infty \sum_{g(x)=a}\frac{f(a)}{|g’(x)|} da=\int_{-\infty}^\infty \sum_{x=a_n}\frac{f(a)}{|g’(x)|}da=\int_{-\infty}^\infty \sum_n\frac{f(a)}{|g’(a_n)|}da=\left(\frac1{|g’(a_1)|}+\dots+\frac1{|g’(a_j)|}\right)\int_{-\infty}^\infty f(a)da$$

Is this correct or how do you write out the sum?

Answer 1

The set $S_a=\{x:g(x)=a\}$ is dependent on $a$. This is the set over which to sum $\frac1{|g'(x)|}$. That is, $$ h(a)=\sum_{g(x)=a}\frac1{|g'(x)|}=\sum_{x\in S_a}\frac1{|g'(x)|} $$ is dependent on $a$, and it cannot be pulled outside of the integral. That is, the integral is $$ \int_{-\infty}^\infty f(g(x))\,\mathrm{d}x=\int_{-\infty}^\infty\sum_{g(x)=a}\frac{f(a)}{|g'(x)|}\,\mathrm{d}a=\int_{-\infty}^\infty h(a)\,f(a)\,\mathrm{d}a $$