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

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 $u=y-x/y$.

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 $f'(x)=-2f(x)$.
A: This is a very unusual problem indeed, especially since solving with a differential equation would probably be overkill. One easy solution is to do the following:
$$ \int_{0}^{\infty} e^{-y^{2}-\left(\dfrac{x}{y}\right)^{2}} dy = \int_{0}^{\infty} e^{-\left(y^{2}+\left(\dfrac{x}{y}\right)^{2}\right)} dy $$
$$ \int_{0}^{\infty} e^{-\left(y^{2}+\left(\dfrac{x}{y}\right)^{2}\right)} dy = \int_{0}^{\infty} e^{-\left(y^{2}-2x+\left(\dfrac{x}{y}\right)^{2}+2x\right)} dy $$
$$ \int_{0}^{\infty} e^{-\left(y^{2}-2x+\left(\dfrac{x}{y}\right)^{2}+2x\right)} dy = \int_{0}^{\infty} e^{-2x-\left(y^{2}-2x+\left(\dfrac{x}{y}\right)^{2}\right)} dy $$
And since we are integrating with respect to y only...
$$ \int_{0}^{\infty} e^{-2x-\left(y^{2}-2x+\left(\dfrac{x}{y}\right)^{2}\right)} dy = e^{-2x} \int_{0}^{\infty} e^{-\left(y^{2}-2x+\left(\dfrac{x}{y}\right)^{2}\right)} dy $$
$$ e^{-2x} \int_{0}^{\infty} e^{-\left(y^{2}-2x+\left(\dfrac{x}{y}\right)^{2}\right)} dy = e^{-2x} \int_{0}^{\infty} e^{-\left(y-\dfrac{x}{y}\right)^{2}} dy $$
Now I'm gonna write the whole thing a bit weirdly, but you'll understand where I'm going in a second:
$$ e^{-2x} \int_{0}^{\infty} e^{-\left(y-\dfrac{x}{y}\right)^{2}} dy = e^{-2x} \int_{0}^{\infty} e^{-\left(1\cdot y-xy^{-1}\right)^{2}} dy $$
According to the Cauchy-Schlömilch transformation, we can assume the following:
$ \int_{0}^{\infty} f(\left(ax-bx^{-1}\right)^{2}) dx = \dfrac{1}{a} \int_{0}^{\infty} f(x^{2}) dx $ for a,b>0
If we take x to be a positive coefficient, we get the following result:
$$ e^{-2x} \int_{0}^{\infty} e^{-\left(1\cdot y-xy^{-1}\right)^{2}} dy = e^{-2x} \int_{0}^{\infty} e^{-y^{2}} dy $$
$$ = \dfrac{\sqrt{\pi}}{2} e^{-2x} $$
And not a single diff eq in sight...
In all honesty, it is unclear to me how one would tackle this problem using only diff eqs instead of some of the more conventional techniques of integration.
A: As a differential equation, the intention was probably to notice that
$$f'(x) = -2\int_0^\infty \frac{x}{y^2}e^{-y^2-\frac{x^2}{y^2}}\:dy = -2\int_0^\infty e^{-\frac{x^2}{y^2}-y^2}\:dy = -2f(x)$$
under the variable interchange $y\leftrightarrow \frac{x}{y}$. Thus we have reduced the problem to the initial value problem
$$\begin{cases}f'+2f=0 \\ f(0) = \int_0^\infty e^{-y^2-\frac{0^2}{y^2}}\:dy = \frac{\sqrt{\pi}}{2}\end{cases}$$
