Difference between revisions of "Lambert transform"
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− | <TR><TD valign="top">[1]</TD> <TD valign="top"> D.V. Widder, "An inversion of the Lambert transform" ''Math. Mag.'' , '''23''' (1950) pp. 171–182</TD></TR> | + | <TR><TD valign="top">[1]</TD> <TD valign="top"> D.V. Widder, "An inversion of the Lambert transform" ''Math. Mag.'' , '''23''' (1950) pp. 171–182 {{DOI|10.2307/3029825}} {{ZBL|0036.35302}}</TD></TR> |
− | <TR><TD valign="top">[2]</TD> <TD valign="top"> V.A. Ditkin, A.P. Prudnikov, "Integral transforms" ''Progress in Math.'' , '''4''' (1969) pp. 1–85 ''Itogi Nauk. Mat. Anal. 1966'' (1967) pp. 7–82</TD></TR> | + | <TR><TD valign="top">[2]</TD> <TD valign="top"> V.A. Ditkin, A.P. Prudnikov, "Integral transforms" ''Progress in Math.'' , '''4''' (1969) pp. 1–85 ''Itogi Nauk. Mat. Anal. 1966'' (1967) pp. 7–82 {{ZBL|0197.37903}}</TD></TR> |
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Latest revision as of 21:25, 4 October 2017
The integral transform $$ F(x) = \int_0^\infty \frac{t a(t)}{e^{xt}-1} dt \ . $$
The Lambert transform is the continuous analogue of the Lambert series (under the correspondence $t a(t) \leftrightarrow a_n$, $e^x \leftrightarrow 1/t$). The following inversion formula holds: Suppose that $$ a(t) \in L(0,\infty) $$ and that $$ \lim_{t \rightarrow +0} a(t) t^{1-\delta} = 0,\ \ \delta > 0 \ . $$
If also $\tau > 0$ and if the function $a(t)$ is continuous at $t = \tau$, then one has $$ \tau a(\tau) = \lim_{k\rightarrow\infty} \frac{(-1)^k}{k!} \left({\frac{k}{\tau}}\right)^{k+1} \sum_{n=1}^\infty \mu(n) n^k F^{(k)}\left({ \frac{nk}{\tau} }\right) \,, $$ where $\mu(n)$ is the Möbius function.
References
[1] | D.V. Widder, "An inversion of the Lambert transform" Math. Mag. , 23 (1950) pp. 171–182 DOI 10.2307/3029825 Zbl 0036.35302 |
[2] | V.A. Ditkin, A.P. Prudnikov, "Integral transforms" Progress in Math. , 4 (1969) pp. 1–85 Itogi Nauk. Mat. Anal. 1966 (1967) pp. 7–82 Zbl 0197.37903 |
Lambert transform. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Lambert_transform&oldid=42007