3
$\begingroup$

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.
$\endgroup$
1
$\begingroup$

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

$$I=\frac{1}{2}$$

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.