If int_{-infty}^{infty} f(x) dx = 1,then int_ | vedclass

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(B) Let $I = \int_{-\infty}^{\infty} f\left(x - \frac{1}{x}\right) dx$. We use the property $\int_{-\infty}^{\infty} g\left(x - \frac{1}{x}\right) dx

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$1$

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(B) Let $I = \int_{-\infty}^{\infty} f\left(x - \frac{1}{x}\right) dx$. We use the property $\int_{-\infty}^{\infty} g\left(x - \frac{1}{x}\right) dx = \int_{-\infty}^{\infty} g(u) du$ for a suitable transformation. Let $u = x - \frac{1}{x}$. Then $du = (1 + \frac{1}{x^2}) dx$. This integral is a standard result in definite integration involving the substitution $x - \frac{1}{x} = u$. Specifically,$\int_{-\infty}^{\infty} f\left(x - \frac{1}{x}\right) dx = \int_{-\infty}^{\infty} f(u) du$. Given $\int_{-\infty}^{\infty} f(x) dx = 1$,it follows that $I = 1$.

If $\int_{-\infty}^{\infty} f(x) dx = 1$,then $\int_{-\infty}^{\infty} f\left(x - \frac{1}{x}\right) dx$ is equal to

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