A Fascinating Integral

Differentiating the expression $\int_{1}^{xy} f(t) dt = y\int_{1}^{x}f(t)dt + x\int_{1}^{y}f(t)dt$ wrt $x$ on both sides using Newton-Leibniz formula, we get $yf(xy) = yf(x) + \int_{1}^y f(t) dt$

Put $x = 1$ , we get,

$yf(y) = y + \int_{1}^y f(t) dt$ $\hspace{5mm} (\because f(1) = 1)$

Again differentiating the above expression wrt $y$ (and using Newton-Leibniz again),

$f(y) + yf'(y) = 1 + f(y)$

$\implies yf'(y) = 1$

$\implies f'(y) = \frac{1}{y}$

Integrating on both the sides we get,

$f(y) = log(y) + C$

Since $f(1) = 1$ ,

$f(y) = log(y) + 1$

Substituting this into $g(x)$ we get,

$g(x) = -({x}^2 + \frac{1}{x^2})$

So now, we have to find $I$ where,

$I = \int_{0}^\infty e^{-(x^2 + \frac{1}{x^2})} dx$

Writing $x \rightarrow \frac{1}{x}$ ,

$I = \int_{0}^\infty e^{-(x^2 + \frac{1}{x^2})} \frac{1}{x^2} dx$

On adding these two equations,

$2I = \int_{0}^\infty e^{-(x^2 + \frac{1}{x^2})} (1 + \frac{1}{x^2}) dx$

Writing $({x}^2 + \frac{1}{x^2}) = (x - \frac{1}{x})^2 + 2$ ,

$2I = \int_{0}^\infty e^{-(x - \frac{1}{x})^2 - 2} (1 + \frac{1}{x^2}) dx$

$\implies 2{e}^2I = \int_{0}^\infty e^{-(x - \frac{1}{x})^2} (1 + \frac{1}{x^2}) dx$

Substitute, $(x - \frac{1}{x}) \rightarrow t$ and changing the limits accordingly, we get,

$2{e}^2I = \int_{-\infty}^{\infty} e^{-t^2} dt$

$\int_{-\infty}^{\infty} e^{-t^2} dt$ is the famous Gaussian Integral,

$\int_{-\infty}^{\infty} e^{-t^2} dt = \sqrt\pi$

Using this we get that,

$I = \frac{\sqrt\pi}{2e^2} \approx 0.119937....$