Calculate the integral connected with $e^{-x^2}$

Question

$$\int_{-\infty}^{\infty} e^{-\frac{1}{2}(\frac{t^2}{y^2}+y^2)}\mathrm{d}y.$$ My initial idea was to present the expression in paranthesis as some square, for instance ($\frac{t}{y}+y)^2$ and exclude the part with t to the front of the integral, then substitute $\frac{t}{y}+y$ with some z, but it does not work, since after differentiating $\frac{t}{y}+y$ we obtain $-y^2$ in the denominator.

Answer 1

Let's call our function $f(t)$:

$$f(t)=\int_0^\infty e^{-\frac12 \left(y^2+\frac{t^2}{y^2} \right)} dy$$

Changing the variable $y \to 1/y$ bring us:

$$f(t)=\int_0^\infty e^{-\frac12 \left(t^2y^2+\frac{1}{y^2} \right)} \frac{dy}{y^2} $$

Now let's find the derivative:

$$f'(t)=-t\int_0^\infty e^{-\frac12 \left(t^2y^2+\frac{1}{y^2} \right)} dy $$

Changing the variable $y \to y/t$ we get:

$$f'(t)=-f(t)$$

The solution is:

$$f(t)=Ae^{-t}$$

The constant is easily found from the original integral for $t=0$.