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Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)

Solving integral equations by $ L$ and $ L\sp{-1}$ operators


Author: Charles Fox
Journal: Proc. Amer. Math. Soc. 29 (1971), 299-306
MSC: Primary 44.28; Secondary 45.00
MathSciNet review: 0280944
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Abstract:

$\displaystyle g(u) = \int_0^\infty {k(ux)f(x)dx = \frac{1}{{2\pi i}}\int_C^{} {K(s)F(1 - s){u^{ - s}}ds,} }$ ($ 1$)

where $ g(u)$ and $ k(u)$ are known and $ f(x)$ is to be found. $ K(s)$ is the Mellin transform of $ k(x)$ and $ F(s)$ of $ f(x)$; hence the second equality. L and $ {L^{ - 1}}$ denote the Laplace transform and its inverse. If

$\displaystyle {{K(s) = \prod\limits_{i = 1}^n {\Gamma ({\alpha _i}s + {\beta _i})} } {\prod\limits_{j = 1}^m {\Gamma ({\alpha _j}s + {\beta _j})} }}$ ($ 2$)

then I show that a suitable combination of L and $ {L^{ - 1}}$ operators, applied to (1), can eliminate $ K(s)$ from the second integrand. This leaves $ F(1 - s)$ standing free and the Mellin transform then obtains $ f(x)$ from $ F(1 - s)$. This solution needs tables of Laplace transforms only.

When (2) does not hold, an L and $ {L^{ - 1}}$ combination may turn (1) into an integral equation whose solution is already known.


References [Enhancements On Off] (What's this?)

  • [1] Tables of integral transforms. Vol. 1, McGraw-Hill, New York, 1954.
  • [2] Tables of integral transforms. Vol. 2, McGraw-Hill, New York, 1954.
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  • [4] E. C. Titchmarsh, Theory of Fourier integrals, Clarendon Press, Oxford, 1937.
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Additional Information

DOI: http://dx.doi.org/10.1090/S0002-9939-1971-0280944-6
PII: S 0002-9939(1971)0280944-6
Keywords: Mellin type integrals and integrands, powers of L and $ {L^{ - 1}}$, Laplace transform, Gamma function factors, integral equations
Article copyright: © Copyright 1971 American Mathematical Society