Abstract
You should always be alert, when confronted by a definite integral, for the happy possibility that although the integral might look ‘interesting’ (that is, hard!) just maybe it will still yield to a direct, frontal attack. The first six integrals in this chapter are in that category. If a and b are positive constants, calculate:
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Notes
- 1.
In this derivation we’ve assumed \( { \lim}_{\mathrm{x}\to 0}\mathrm{x} \ln \left(1+\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}\right)={ \lim}_{\mathrm{x}\to \infty}\mathrm{x} \ln \left(1+\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}\right)=0 \). To see that these two assumptions are correct, recall the power series expansion for the log function when p ≈ 0: ln(1 + p) = \( \mathrm{p}-\frac{1}{2}{\mathrm{p}}^2+\frac{1}{3}{\mathrm{p}}^3-\cdots \). So, with \( \mathrm{p}=\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2} \) (which → 0 as x → ∞), we have \( \mathrm{x} \ln \left(1+\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}\right)=\mathrm{x}\left[\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}-\frac{1}{2}{\left(\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}\right)}^2+\frac{1}{3}{\left(\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}\right)}^3-\cdots \right]=\frac{{\mathrm{a}}^2}{\mathrm{x}}-\frac{1}{2}\frac{{\mathrm{a}}^4}{{\mathrm{x}}^3}+\cdots \) which → 0 as x → ∞. On the other hand, as x → 0 we have \( \mathrm{x} \ln \left(1+\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}\right)\approx \mathrm{x} \ln \left(\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}\right)=\mathrm{x} \ln \left({\mathrm{a}}^2\right)-\mathrm{x} \ln \left({\mathrm{x}}^2\right)=\mathrm{x} \ln \left({\mathrm{a}}^2\right)-2\mathrm{x} \ln \left(\mathrm{x}\right) \) and both of these terms go to zero as x goes to zero (the first term is obvious, and in the second term x vanishes faster than ln(x) blows-up).
- 2.
Writing the partial fraction expansion this way is where the assumption that all the ai are different comes into play. If any of the ai appears multiple times, then the correct partial fraction expansion of the integrand is not as I’ve written it.
- 3.
There are two points to be clear on at this point. First, since we are working with an identity it must be true for all values of x, and I’ve just picked a particularly convenient one. Second, if the use of an imaginary x bothers you, just remember the philosophical spirit of this book—anything (well, almost anything) goes, and we’ll check our result when we get to the end!
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Nahin, P.J. (2015). ‘Easy’ Integrals. In: Inside Interesting Integrals. Undergraduate Lecture Notes in Physics. Springer, New York, NY. https://doi.org/10.1007/978-1-4939-1277-3_2
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