Integral with variable change

Question

How do I go about solving this

I put $x=t-\frac{1}{t}$ but I'm not able to proceed ahead .

Answer 1

I will use two alternative methods to show that $$\int_{-\infty}^\infty f(x) \, dx = \int_{-\infty}^\infty f \left (x - \frac{1}{x} \right ) \, dx.$$

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Method 1

$$I = \int_{-\infty}^\infty f \left (x - \frac{1}{x} \right ) \, dx = \int_{-\infty}^0 f \left (x - \frac{1}{x} \right ) \, dx + \int_0^\infty f \left (x - \frac{1}{x} \right ) \, dx. \tag1$$ Enforcing a substitution of $x \mapsto -1/x$ in the above integral yields $$I = \int_0^\infty f \left (x - \frac{1}{x} \right ) \frac{dx}{x^2} + \int_{-\infty}^0 f \left (x - \frac{1}{x} \right ) \frac{dx}{x^2}. \tag2$$ Adding (1) to (2) gives \begin{align*} I &= \frac{1}{2} \int_0^\infty f \left (x - \frac{1}{x} \right ) \left (1 + \frac{1}{x^2} \right ) \, dx + \frac{1}{2} \int^0_{-\infty} f \left (x - \frac{1}{x} \right ) \left (1 + \frac{1}{x^2} \right ) \, dx. \end{align*}

Now let $u = x - \frac{1}{x}, du = (1 + \frac{1}{x^2}) \, dx$, gives $$I = \frac{1}{2} \int_{-\infty}^\infty f(u) \, du + \frac{1}{2} \int_{-\infty}^\infty f(u) \, du,$$ or $$ \int_{-\infty}^\infty f(x) \, dx = \int_{-\infty}^\infty f \left (x - \frac{1}{x} \right ) \, dx,$$ as required.

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Method 2

$$I = \int_{-\infty}^\infty f \left (x - \frac{1}{x} \right ) \, dx = \int_{-\infty}^0 f \left (x - \frac{1}{x} \right ) \, dx + \int_0^\infty f \left (x - \frac{1}{x} \right ) \, dx.$$ In the first integral appearing on the right, let $x = -e^{-u}$. In the second let $x = e^u$. Thus \begin{align*} I &= \int_{-\infty}^\infty f \left (-e^{-u} + \frac{1}{e^{-u}} \right ) e^{-u} \, du + \int_{-\infty}^\infty f \left (e^u - \frac{1}{e^u} \right ) e^u \, du\\ &= \int_{-\infty}^\infty f(e^u - e^{-u})(e^u + e^{-u}) \, du\\ &= \int_{-\infty}^\infty f(2 \sinh u) \cdot 2 \cosh u \, du. \end{align*} Now if we let $x = 2 \sinh u, dx = 2 \cosh u \, du$, giving $$\int_{-\infty}^\infty f \left (x - \frac{1}{x} \right ) \, dx = \int_{-\infty}^\infty f(x) \, dx,$$ as before.

Answer 2

Let $$u=x-\frac{1}{x}$$ Then $$x=\frac{u\pm\sqrt{u^2+4}}{2}$$

The integral becomes $$\int_{-\infty}^0f(x-\frac{1}{x})dx+\int_0^\infty f(x-\frac{1}{x})dx$$ $$=\int_{-\infty}^\infty f(u)d\frac{u-\sqrt{u^2+4}}{2}+\int_{-\infty}^\infty f(u)d\frac{u+\sqrt{u^2+4}}{2}$$ $$=\int_{-\infty}^\infty f(u)d(u)=1$$