# Evaluate the integral $\int_1^\infty \frac{2^x}{2^{(2^x)}}dx$

Evaluate the integral $$\int_1^\infty \frac{2^x}{2^{(2^x)}}dx$$

My Try:

substituting $t = 2^x$ we get:

$$\ln 2 \int_2^\infty \left(\frac{1}{2}\right)^t dt = \frac{\ln 2}{\ln 0.5} \left( \left(\frac{1}{2}\right)^\infty - \left(\frac{1}{2}\right)^2 \right)$$

Apparently $\frac{\ln 2}{\ln 0.5} = -1$ so we get that the integral equals $\frac{1}{4}$.

But that's a false proof.

Where is my mistake and how to correct that?

Thanks.

• What different it would make? $(1/2)^n \to 0$. – Elimination Jan 30 '15 at 15:21
• $ln(2)$ goes into denominator – Arashium Jan 30 '15 at 15:24

$t=2^x$,$x=\frac{\ln t}{\ln 2}$,thus $dx=\frac{1}{\ln2}\frac1tdt$, I'm afraid the integral should be $$\int_1^\infty \frac{2^x}{2^{(2^x)}}dx=\frac{1}{\ln 2} \int_2^\infty \left(\frac{1}{2}\right)^t dt$$
• Yes, and the final answer is $\frac{1}{ln(2) ln(0.5)}\times (0-\frac{1}{4})$ – Arashium Jan 30 '15 at 15:27