Prove that $e$ is the root of the equation $\int_0^{\infty} \frac{dt}{(t+x)\sqrt{4t+(x+1)^2}}=\frac{1}{x-1}$

Question

It seems numerically that $e$ is the only real root of the equation:

$$\int_0^{\infty} \frac{dt}{(t+x)\sqrt{4t+(x+1)^2}}=\frac{1}{x-1}$$

Mathematica confirms it at least to the large number of digits:

$$\int_0^{\infty} \frac{dt}{(t+e)\sqrt{4t+(e+1)^2}}=\frac{1}{e-1}=0.5819767068693$$

I stumbled on this more or less accidentally, so I have two questions:

How can we prove that $e$ is the only root of this equation?

Can we create some kind of iteration algorithm to compute $e$ this way?

Edit

By the way, the second question can be answered positively, even though numerical integration is involved. We can just use the fixed point iterations:

$$I_n=\int_0^{\infty} \frac{dt}{(t+x_n)\sqrt{4t+(x_n+1)^2}}$$

$$x_{n+1}=1+\frac{1}{I_n}$$

We can take something like $2$ as initial guess.

So, the second question was more about- can we do that without direct numerical integration?

Answer 1

There is a primitive of the integral $$ \int\frac{dt}{(t+x)\sqrt{4t+(x+1)^2}}=\frac{\log \left(\sqrt{4 t+(x+1)^2}-x+1\right)-\log \left(-\sqrt{4 t+(x+1)^2}-x+1\right)}{x-1}\ . $$ Therefore for $x>0$ the definite integral reads $$ \int_0^{\infty} \frac{dt}{(t+x)\sqrt{4t+(x+1)^2}}=\frac{\log x}{x-1}\ . $$ Therefore for $x>0$ you only have to solve $$ \frac{\log x}{x-1}=\frac{1}{x-1}\Rightarrow x=\mathrm{e}\ , $$ which is the only solution. For $x<0$, your integral is only defined as a principal value integral.

Answer 2

Can we create some kind of iteration algorithm to compute e this way?

Yes, you can use any root solving algorithm.