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 enter image description here

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

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


  • $\begingroup$ Your $I(t)$ look suspiciously divergent at $y=0$ (which means that your splitting on the line above splits the integral into the difference of two divergent integrals). $\endgroup$ – mickep Nov 23 '15 at 13:55
  • $\begingroup$ @mickep do you mean to say $I(t)$ is divergent at $t=0$, i could not get in saying divergent at $y=0$ $\endgroup$ – Ekaveera Kumar Sharma Nov 23 '15 at 14:11
  • $\begingroup$ No, not $t=0$. For example, the integral $\int_0^{+\infty} 2^{-x}/x$ is divergent. The problem there is at $x=0$, since there your function behaves like $2^{-x}/x\approx 1/x$, and the integral of $1/x$ is divergent (when the lower limit is zero). $\endgroup$ – mickep Nov 23 '15 at 15:09

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 $$I=\frac{\text{Ei}\left(-2^{2 t} \log (2)\right)-\text{Ei}\left(-2^{2 t+1} \log (2)\right)}{2 \log (2)}$$


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