How do I calculate

$$u(w)=\int_0^\infty \frac{1-\cos(wt)}{t}\,e^{-t}\,dt$$

I tried to do it, I use partial integration but I get lost. Is there any nice simple way to calculate it?

  • $\begingroup$ There is a missing left parenthesis. $\endgroup$ – Jean Marie Aug 26 '16 at 21:08
  • 1
    $\begingroup$ Please put the left parenthesis where it is supposed to be and clarify that $u$ has to be intended as a function of the $w$ variable. $\endgroup$ – Jack D'Aurizio Aug 26 '16 at 21:19
  • $\begingroup$ Note that $t$ is only a dummy variable in the integral. Is $w$ (or $\omega$) a parameter or a constant here? $\endgroup$ – Brian Tung Aug 26 '16 at 21:20
  • $\begingroup$ I guess the actual question is to compute $$\frac{d}{dw}\int_{0}^{+\infty}\frac{1-\cos(wt)}{t}e^{-t}\,dt. $$ $\endgroup$ – Jack D'Aurizio Aug 26 '16 at 21:24

For any $w\in\mathbb{R}$, the identity $$ \int_{0}^{+\infty}\frac{1-\cos(wt)}{t}e^{-t}\,dt = \frac{1}{2}\log(1+w^2) \tag{1} $$ follows from the complex version of Frullani's theorem, hence $$ \frac{d}{dw}\int_{0}^{+\infty}\frac{1-\cos(wt)}{t}e^{-t}\,dt = \frac{w}{1+w^2}.\tag{2} $$ You may prove the same by checking first that, by the dominated convergence theorem, we may apply differentiation under the integral sign, leaving us with $$ \frac{d}{dw}\int_{0}^{+\infty}\frac{1-\cos(wt)}{t}e^{-t}\,dt =\int_{0}^{+\infty}t \sin(wt) e^{-t}\,dt \tag{3}$$ where the last integral is easy to compute by integration by parts.


$\int \frac {(1−\cos\omega t)e^{−t}}{t}dt$ cannot be evaluated into elementary functions. You need to get tricky.

$F(s) = \int_0^{\infty} \frac {(1−\cos\omega t)e^{−st}}{t}dt$ and if we can find $F(1)$ we are done.

$\frac {dF}{ds} = $$\int_0^{\infty} -(1−\cos\omega t)e^{−st}dt\\ \frac 1s e^{-st} + \frac {-s\cos\omega te^{−st} + \omega \sin\omega t e^{−st}}{s^2+\omega^2} |_0^\infty\\ -\frac 1s + \frac {s}{s^2+\omega^2}$

$F(\infty) - F(1) = \int_1^{\infty}-\frac 1s + \frac {s}{s^2+\omega^2}\\ F(\infty) - F(1) = -\ln s + \frac 12 (s^2 + \omega^2)|_1^{\infty}$

I am going to leave it to you to prove to yourself that $\lim_\limits{s\to\infty} \ln s + \frac 12 (s^2 + \omega^2) = 0$

$F(\infty) - F(1) = -\frac 12 \ln(1 + \omega^2)$

Going back to the definition of $F$, it should be clear that $F(\infty) = 0$

$F(1) = \frac 12 \ln(1 + \omega^2)$


Here is an approach that does not use Feynman's trick or the Generalized Frullani's Theorem, but rather uses integration by parts ("IBP") along with the well-known integral (See the NOTE at the end of THIS ANSWER$$\int_0^\infty \log(t)e^{-t}\,dt=-\gamma\tag1$$where $\gamma$ is the Euler-Mascheroni constant. It is easy to see from $(1)$ that for $\text{Re}(s)>0$ $$\int_0^\infty \log(t)e^{-st}\,dt=-\frac{\gamma-\log(s)}s\tag2$$

First, by enforcing the substitution $x= t/w$ reveals that

$$\int_0^\infty \frac{1-\cos(wx)}{x}e^{-x}\,dx=\int_0^\infty \frac{1-\cos(t)}{t}e^{-(1/w)t}\,dt$$

Letting $s=1/w$, we have

$$\begin{align} \int_0^{\infty} \left(\frac{1- \cos t}{t}\right)e^{-st}dt&=\int_0^{\infty} (1- \cos t)\left(\frac{d\log(t)}{dt}\right)e^{-st}dt\\\\ &=\int_0^{\infty} \left(e^{-st}- \frac12 e^{-(s+i)t}-\frac12 e^{-(s-i)t}\right)\left(\frac{d\log(t)}{dt}\right)\,dt\\\\ &\overbrace{=}^{IBP}-\int_0^{\infty} \log(t)\frac{d}{dt}\left(e^{-st}- \frac12 e^{-(s+i)t}-\frac12 e^{-(s-i)t}\right) dt\\\\ &=s\int_0^{\infty} \log(t)e^{-st}dt\\\\ &-\frac12(s+i)\int_0^{\infty} \log(t)e^{-(s+i)t}dt\\\\ &-\frac12(s-i)\int_0^{\infty} \log(t)e^{-(s-i)t}dt\\\\ &\overbrace{=}^{\text{Using} (2)}s\left(\frac{-\gamma-\log(s)}{s}\right)\\\\ &-\frac12 (s+i)\left(\frac{-\gamma-\log(s+i)}{s+i}\right)\\\\&-\frac12 (s-i)\left(\frac{-\gamma-\log(s-i)}{s-i}\right)\\\\ &=\frac12 \log\left(\frac{s^2+1}{s^2}\right) \end{align}$$

Finally, replacing $s$ with $1/w$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{1-\cos(wx)}{x}e^{-x}\,dx=\frac12\log\left(1+w^2\right)}$$


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