If $ \int_{-\infty}^{\infty} f(x) dx= 1,$ Find value of $ \int_{-\infty}^{\infty} f(x-\frac{1}{x}) dx$.
I tried substituting $x$ as $\frac{1}{t}$, but nothing is happening. In the denominator, $1+x^2$ is left.
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Hint: Try substituting $x=e^u$. – John Doe Jan 05 '19 at 17:49
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@JohnDoe Your substitution only works for $x\in(0,\infty)$. – Matemáticos Chibchas Jan 13 '19 at 06:01
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@matematicoschibchas hmm, you're right. You can split it the integral range into positive and negative $x$, then use $x=-e^{-u}$ for the negative part. – John Doe Jan 14 '19 at 02:20
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One may recall the following property $$ \int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{+\infty} f\left( x - \frac{1}{x} \right) \, dx $$ proved here.
Olivier Oloa
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