Find the derivative of the function $f(x)=\int_{0}^\infty e^{-y^2-(x/y)^2} dy$

Question

This problem confuses me, Can we use Leibniz rule here? But in the chapter Leibniz rule was not introduced. Is it possible to find the derivative using simple concepts. The actual problem is to evaluate the integral $\int_{0}^\infty e^{-y^2-(9/y)^2} dy$ using differential equations.

Answer 1

This is the correct first step in evaluating $\int_0^\infty e^{-y^2-(9/y)^2}\,\mathrm{d}y$.

To justify the swapping of the order of integration and differentiation (differentiation inside the integral), we can use Fubini and FTC: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^\infty e^{-y^2-x^2/y^2}\,\mathrm{d}y &=\frac{\mathrm{d}}{\mathrm{d}x}\int_0^\infty\int_x^\infty\frac{2t}{y^2}e^{-y^2-t^2/y^2}\,\mathrm{d}t\,\mathrm{d}y\tag{1a}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\int_x^\infty\int_0^\infty\frac{2t}{y^2}e^{-y^2-t^2/y^2}\,\mathrm{d}y\,\mathrm{d}t\tag{1b}\\ &=-\int_0^\infty\frac{2x}{y^2}e^{-y^2-x^2/y^2}\,\mathrm{d}y\tag{1c} \end{align} $$ Explanation:
$\text{(1a)}$: $\int_x^\infty\frac{2t}{y^2}e^{-t^2/y^2}\,\mathrm{d}t=e^{-x^2/y^2}$
$\phantom{\text{(1a):}}$ here we are writing a function
$\phantom{\text{(1a):}}$ as the integral of its derivative
$\text{(1b)}$: Fubini-Tonelli
$\text{(1c)}$: Fundamental Theorem of Calculus

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Hint 1:

For the next step, you might want to complete the square in the exponents $$ f(x)=\int_0^\infty e^{-y^2-x^2/y^2}\,\mathrm{d}y=e^{-2x}\int_0^\infty e^{-(y-x/y)^2}\,\mathrm{d}y\tag2 $$ and from $(1)$ $$ -\frac12f'(x)=\int_0^\infty\frac{x}{y^2}\,e^{-y^2-x^2/y^2}\,\mathrm{d}y=e^{-2x}\int_0^\infty\frac{x}{y^2}e^{-(y-x/y)^2}\,\mathrm{d}y\tag3 $$ then consider the substitution $u=y-x/y$.

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Hint 2:

Another approach to the next step was suggested by Sangchul Lee: note that $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^\infty e^{-y^2-x^2/y^2}\,\mathrm{d}y &=-\int_0^\infty\frac{2x}{y^2}e^{-y^2-x^2/y^2}\,\mathrm{d}y\tag{4a}\\ &=-2\int_0^\infty e^{-y^2-x^2/y^2}\,\mathrm{d}y\tag{4b}\\ \end{align} $$ Explanation:
$\text{(4a)}$: $(1)$ allows us to differentiate inside the integral
$\text{(4b)}$: substitute $y\mapsto x/y$

and $(4)$ says that $f'(x)=-2f(x)$.