# $\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ [duplicate]

I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead.

$$\int\sqrt{x\cdot \sqrt{x\cdot \sqrt{x\cdot\sqrt{x\ldots } }}}\,dx\tag{1}$$

$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx\tag{2}$$

I was able to solve (1), as the integrand simplifies to $$x^{e-2}$$, however, I'm struggling with solving (2).

If we rewrite the roots as powers, we get:

$$\int x^\frac{2}{2}\cdot x^\frac{3}{4}\cdot x^\frac{4}{8}\cdot x^\frac{5}{16}\ldots\,dx$$

combining the powers we get: $$\int x^{\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\ldots}$$

the exponent is the infinite sum

$$\sum^{\infty}_{n=1}\frac{n+1}{2^n}\tag{3}$$ we can split this into: $$\sum^{\infty}_{n=1}\frac{n}{2^n}+\sum^{\infty}_{n=1}\frac{1}{2^n}$$ The right sum is well known except here the sum begins at $$n=1$$, meaning that the right sum evaluates to 1. Messing around with desmos, the integrand appears to be $$x^3,x>0$$ implying that (3) converges to 3 and the $$\sum^{\infty}_{n=1}\frac{n}{2^n}$$ converges to 2.

Which is part I'm struggling with. Any ideas?

$$\sum^{\infty}_{n=1}\frac{n}{2^n}$$

Consider $$\sum_{n=1}^\infty n x^n=x\sum_{n=1}^\infty n x^{n-1}=x\left(\sum_{n=1}^\infty x^{n}\right)'$$
When finished, make $$x=\frac 12$$
If $$f(x)=\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$$ $$\forall$$ $$\vert x\vert \lt 1$$. Then what is $$\frac{f'\left(\frac{1}{2}\right)}{2}$$
The binomial theoem gives$$(1-x)^{-2}=\sum_{n\ge0}(n+1)(-1)^nx^n,$$so the exponent is$$-1+(1-1/2)^{-2}=3.$$So the integral is $$\int x^3dx=\frac14x^4+C$$.