# Clarify and justify how get the derivative of the Laplace transform of the Buchstab function

I would like to justify that the derivative with respect to $s$ of the Laplace transform of the Buchstab function is $$\int_1^\infty u\omega(u)e^{-su}du=\frac{e^{-s}}{s}\exp\left(\int_0^\infty \frac{e^{-t}}{t}dt\right)$$

for $s>0$.

Question. Was my deduction right? Please can you justify it rigorously? Thanks in advance.

You can take the definiton of Buchstab function, and its Laplace transform from Tao What's new? I say Exercise 28 of 254A, Supplement 4: Probabilistic models and heuristics for primes, and take required theorems from Wikipedia Differentiation under the integral sign. I¡ve computed the derivative

$$\frac{d}{ds}\int_0^\infty \frac{e^{-t}}{t}dt$$ with Wolfram Alpha because I had doubts of how work with the upper limit, since there is infinite (also with calculus, I say the rule chain, and LHS by other theorem). I need these calculations since I would like to study more calculations for this function, like to try the integration by parts (notice that it is possible a simplification if one uses the related differential equation). Then can you help to justify the right statement, I say both sides of previous derivative of the Laplace transform of this special function with respect $s$. Also if you want clarify, if in LHS the interval of integration is $\int_0^\infty$ or $\int_1^\infty$

The Laplace transform that you are looking for is given, of course, by the formula

$$\mathcal L (\omega) (s) = \int \limits _0 ^\infty \omega (u) \ \Bbb e ^{-su} \ \Bbb d u .$$

The problem is that the definition of Buchstab's function (the series in Tao's Ex. 28.i) is almost useless for computations. Fortunately, there comes Ex. 28.iii which gives the following formula:

$$u \omega (u) = 1_{(1,\infty)} (u) + \int \limits _0 ^u 1_{(1,\infty)} (t) \ \omega (u-t) \ \Bbb d t .$$

We have, therefore, to produce an $u \omega (u)$ inside Laplace's transform, and the obvious way to to this is to derive with respect to $s$:

$$\mathcal L (\omega) ' (s) = \int \limits _0 ^\infty -u \omega (u) \ \Bbb e ^{-su} \ \Bbb d u = - \int \limits _0 ^\infty \left( 1_{(1,\infty)} (u) + \int \limits _0 ^u 1_{(1,\infty)} (t) \ \omega (u-t) \ \Bbb d t \right) \ \Bbb e ^{-su} \ \Bbb d u = \\ - \int \limits _1 ^\infty \Bbb e ^{-su} \ \Bbb d u - \int \limits _0 ^\infty \left( \int \limits _0 ^u 1_{(1,\infty)} (t) \ \omega (u-t) \ \Bbb d t \right) \ \Bbb e ^{-su} \ \Bbb d u = \\ - \frac {\Bbb e ^{-s}} s - \int \limits _1 ^\infty \left( \int \limits _1 ^u \omega (u-t) \ \Bbb d t \right) \ \Bbb e ^{-su} \ \Bbb d u = - \frac {\Bbb e ^{-s}} s - \int \limits _1 ^\infty \left( \int \limits _t ^\infty \omega (u-t) \ \Bbb d u \right) \ \Bbb e ^{-su} \ \Bbb d t = \\ - \frac {\Bbb e ^{-s}} s - \int \limits _1 ^\infty \left( \int \limits _0 ^\infty \omega (v) \ \Bbb d v \right) \ \Bbb e ^{-s(t+v)} \ \Bbb d t = - \frac {\Bbb e ^{-s}} s - \int \limits _1 ^\infty \Bbb e ^{-st} \ \Bbb d t \ \mathcal L (\omega) (s) = \\ - \frac {\Bbb e ^{-s}} s \big( \mathcal L (\omega) (s) + 1 \big) .$$

This is an elementary differential equation which can be rewritten as

$$\frac {\mathcal L (\omega) ' (s)} {\mathcal L (\omega) (s) + 1} = - \frac {\Bbb e ^{-s}} s ,$$

whence it follows that, for arbitrary but fixed $\sigma > 0$,

$$\ln \big( \mathcal L (\omega) (s) + 1 \big) - \ln \big( \mathcal L (\omega) (\sigma) + 1 \big) = - \int \limits _\sigma ^s \frac {\Bbb e ^{-t}} t \ \Bbb d t$$

so finally, after exponentiating and rearranging a bit,

$$\mathcal L (\omega) (s) = \exp \left( - \int \limits _\sigma ^s \frac {\Bbb e ^{-t}} t \ \Bbb d t \right) \big( \mathcal L (\omega) (\sigma) + 1 \big) - 1 .$$

It is now easy to derive with respect to $s$ and get

$$\mathcal L (\omega) ' (s) = - \frac {\Bbb e ^{-s}} s \ \exp \left( - \int \limits _\sigma ^s \frac {\Bbb e ^{-t}} t \ \Bbb d t \right) \big( \mathcal L (\omega) (\sigma) + 1 \big) .$$

Your own result is very close to being correct, having the following errors:

• the minus sign missing from the fraction (both inside and outside the exponential)

• the endpoints of your integral in the RHS are $0$ and $\infty$; $0$ clearly cannot be, because $\frac 1 s$ is not integrable close to $s=0$ (the numerator $\Bbb e ^{-s}$ not raising any problem)

• the factor $\mathcal L (\omega) (\sigma) + 1$ has to be there, playing the role of the initial condition for that differential equation; unfortunately, with $\omega$ being so ugly, there is no obvious $\sigma$ for which $\mathcal L (\omega) (\sigma)$ could have a simple value, therefore we have to leave it there.

• Very thanks much, this night I will study your remarks in this nice answer. Very thanks much.
– user243301
Jul 26, 2016 at 12:46
• There was a lot of merit from you to get these nice calculations, I read the answer and understand many calculations, I hope in the next future reading this to obtain your abilities. Thanks.
– user243301
Jul 26, 2016 at 19:11