An almost Fresnel integral

Hint. One may perform the change of variable $$ u=2^x, \quad \ln x= \frac1{\ln 2}\cdot \ln u, \quad dx=\frac1{\ln 2}\cdot \frac{du}u, $$ giving $$ I=\int_0^{+\infty}\sin(2^x)\ dx=\frac1{\ln 2}\cdot\int_1^{+\infty}\frac{\sin(u)}{u}\ du=\frac1{\ln 2}\cdot\left(\frac{\pi }{2}-\text{Si}(1)\right) $$ where we have made use of the sine integral function $\text{Si}(\cdot)$.

If you perform this change of variable:

$$y = 2^x \Rightarrow \begin{cases} x = \frac{\log(y)}{\log(2)}\\ dx = \frac{dy}{y \log(2)}\\ x = 0 \Rightarrow y = 1\\ x = +\infty \Rightarrow y = +\infty \end{cases},$$

then you obtain the following: $$\int_0^{+\infty} \sin(2^x)dx = \frac{1}{\log(2)} \int_1^{+\infty} \frac{\sin(y)}{y} dy = \frac{\text{Si}(+\infty) - \text{Si}(1)}{\log(2)},$$

where $\text{Si}(x)$ is the sine integral. I guess you can't do more than this.