Prove that this expression is equal to $\pi$

Question

Today, on the auspicious $\pi$ day, I saw on a local chat group $$\pi=4\phi^2\left(\phi^2+2\sqrt{\phi}\right)\left(\int_{\ 0}^{\ \infty\ }e^{x^2}\frac{\ \sin\left(x^2\right)\ }{x^2}dx\right)^2$$ I wonder how one can prove this.

My first thoughts were to evaluate the term involving the golden ratio. I calculated it approximately with the help of wolframaplha but you see that I can't prove it with mathematical methods. And for the integral I've no idea as I don't know how to solve improper integrals.

This is what I saw

Edit: I confirmed this query with the Original poster. Indeed it should be $e^{{-x}^2}.$ Surprised that wolfram got it wrong...

Answer 1

As said in my first comment above, the integral is divergent (so: wolframaplha is wrong). Indeed, let $a_k:=\frac\pi2+2k\pi$ with $\Bbb N\ni k\to+\infty$ then $$\int_{\sqrt{a_k-\frac\pi4}}^{\sqrt{a_k+\frac\pi4}}e^{x^2}\frac{\sin(x^2)}{x^2}dx=\frac12\int_{a_k-\frac\pi4}^{a_k+\frac\pi4}e^y\frac{\sin y}{y^{3/2}}dy $$ $$\ge\frac{e^{a_k-\frac\pi4}}{2\sqrt2(a_k+\frac\pi4)^{3/2}}\to+\infty$$

Answer 2

You are doing very tricky things here, like $\phi^2 + 2\sqrt{\phi}$.
Let's have a look at this:

As $\phi^2 = \phi + 1$, we can say that:

$$\phi^2 + 2\sqrt{\phi}$$ $$ = \phi + 2\sqrt{\phi} + 1$$ $$ = (\sqrt{\phi} + 1)^2$$

In top of that, as mentioned in some comments, you're not dealing with $e^{x^2}$, but with $e^{-x^2}$, so you need to prove:

$$\int_{\ 0}^{\ \infty\ }e^{-x^2}\frac{\ \sin\left(x^2\right)\ }{x^2}dx = \frac {\sqrt{\pi}}{2 \phi \cdot (\sqrt{\phi} + 1)}$$

In case you doubt about the replacement of $e^{x^2}$ by $e^{-x^2}$:

This are the graphs of both (as long as Geogebra does not go nuts on the exponential numbers :-) ):

That should give you something to start with :-)

Answer 3

$$I=\int\frac{e^{-x^2} \sin \left(x^2\right)}{x^2}\,dx=\Im\int\frac{e^{-(1-i) x^2}}{x^2} \,dx=\frac 12 \Im\int \frac{e^{-(1-i) t}}{ t^{3/2}}\,dt$$ One integration by parts leads to an incomplete gamma function and using the bounds $$I=\int\frac{e^{-x^2} \sin \left(x^2\right)}{x^2}\,dx=\sqrt{\frac{1}{2} \left(\sqrt{2}-1\right) \pi }\quad \implies\quad I^2=\frac{\sqrt{2}-1}{2} \pi$$

No way to get it.

Answer 4

For $z\in\mathbb{C}$ with $\Re z>0$ we have $$\int_0^\infty e^{-zt^2}\,dt=\frac{\sqrt\pi}{2\sqrt{z}}\implies\int_0^\infty\frac{1-e^{-zt^2}}{t^2}\,dt=\sqrt{\pi z}.$$ This gives, as an example, $$\int_0^\infty e^{-x^2}\frac{\sin x^2}{x^2}\,dx=\Im\sqrt{\pi(1+i)}=\sqrt{\frac\pi2(\sqrt2-1)}.$$