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Using double integrals to solve single integrals

Published online by Cambridge University Press:  14 June 2016

G. J. O. Jameson
G. J. O. Jameson*
Affiliation:
Dept. of Mathematics and Statistics, Lancaster University, Lancaster LA1 4YF e-mail: g.jameson@lancaster.ac.uk
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Extract

Consider the integral

where b > a > 0. First, let us clarify why it even exists. Of course, convergence at infinity is ensured by the exponential terms, but the integrals of and e–ax/x and e–bx/x, taken separately, are divergent at 0, since these integrands equate asymptotically to 1/x as x → 0. However,

so (e–ax – e–bx)/x tends to the finite limit b – a as x → 0 and there is no problem integrating it on intervals of the form [0, r].

A neat way to evaluate I1 starts by expressing the integrand itself as an integral:

(1)

Inserting this into I1 converts it into a double integral.

Type
Research Article
Information
The Mathematical Gazette , Volume 100 , Issue 548 , July 2016 , pp. 257 - 265
Copyright
Copyright © Mathematical Association 2016 

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