How to compute $\int_{-\infty}^{+\infty} 2^{-4^t}(1-2^{-4^t})\,dt=\frac{1}{2}.$ 
How to compute
  $$\int_{-\infty}^{+\infty} 2^{-4^t}(1-2^{-4^t})\,dt=\frac{1}{2}.$$
  I'm interested in more ways of computing this integral.


My thoughts :
Let $y=4^t$ we got $\displaystyle\frac{1}{\ln 4}\int_0^{\infty}\dfrac{e^{-y\ln2}-e^{-y\ln4}}{y}d y$ and we 've $\displaystyle\int_0^{\infty}\dfrac{e^{-ay}-e^{-by }}{y}dy=\ln\frac{b}{a}$
(by studing the function 
$\displaystyle f(x)=\int_0^{\infty}\dfrac{e^{-y}-e^{-xy}}{y}dy$).
Frullani's Integral



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*I'm interested in more ways of computing this integral.

 A: Use substitution $4^t=x$ we get
$$I=\frac{1}{ln4}\int_{0}^{\infty}\frac{2^{-x}(1-2^{-x})}{x}\:dx$$ $\implies$
$$ln4 \times I=\int_{0}^{\infty}\frac{2^{-x}}{x}-\int_{0}^{\infty}\frac{2^{-2x}}{x} \tag{1}$$
For all $t \in \mathbb{R^+}$ let $$I(t)=\int_{0}^{\infty}\frac{2^{-ty}dy}{y}$$ Differentiating both sides w.r.t t we get
$$I'(t)=-\int_{0}^{\infty}2^{-ty} \:ln2\: dy=\frac{-1}{t}$$ now solving this differential equation we get
$$I(t)=-lnt+c$$ and for $t=1$ we get
$$I(t)=-lnt+I(1)$$ and if $t=2$
$$I(2)=-ln2+I(1)$$ and from $(1)$
$$ln4 \times I=I(1)-I(2)=ln2$$ so
$$I=\frac{1}{2}$$
A: I do not know if this is of any interest to you; so, please forgive me if I am off-topic.
Considering the antiderivative, you properly arrived to $$I=\frac{1}{2\log (2)}\int\dfrac{e^{-y\log(2)}-e^{-y\log(4)}}{y}\,dy$$ Now, consider the general problem of $$J=\int\dfrac{e^{-ay}}{y}\,dy$$ Change variable $ay=z$ which makes  $$J=\int\dfrac{e^{-z}}{z}\,dy=\text{Ei}(-z)$$ where appears the definition of the exponential integral function. So, back to the initial problem $$I=\frac{\text{Ei}\left(-2^{2 t} \log (2)\right)-\text{Ei}\left(-2^{2
   t+1} \log (2)\right)}{2 \log (2)}$$
