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I am wondering what techniques exist for the asymptotic evaluation of integrals. Consider the integral $$ I(\lambda) = \int_1^\lambda dx \sqrt{1-\frac 1 x} = \sqrt \lambda \sqrt{\lambda - 1}- \cosh^{-1} \sqrt \lambda . $$ It is clear from the explicit expression that $I(\lambda) \approx \lambda - \frac 1 2\log(4 \lambda) - \frac 1 2$ for large $\lambda$. Could we have found this asymptotic solution without explicitly evaluating the integral? I am familiar only with the steepest descent method, but it does not appear to be applicable here.

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It's much easier than steepest descent : just analyse the main behaviour of the integrand, guess where the main contribution is coming from and what shape it takes, substract it, and iterate. Call $I_1(\lambda)$ your integral : $$I_1(\lambda) = \int_1^\lambda \sqrt{1-1/x}\ {\rm d}x .$$ The integrand is $\sim 1$ as $x\to\infty$, so that $$ I_1(\lambda) = \lambda - 1 + \underbrace{\int_1^\lambda (\sqrt{1-1/x}-1){\rm d}x}_{I_2(\lambda)}. $$ Then the integrand in $I_2(\lambda)$ is $\sim -1/(2x)$ as $x\to\infty$, so that $$ I_2(\lambda) = -\frac12\int_1^\lambda\frac{{\rm d} x}x + \underbrace{\int_1^\lambda \big(\sqrt{1-1/x}-1+1/(2x)\big){\rm d}x}_{J_3(\lambda)}. $$ The integrand in $J_3$ is $O(1/x^2)$ for large $x$, so $J_3(\lambda)$ converges to a constant as $\lambda\to\infty$. Write $$ J_3(\lambda) = \int_1^\infty \big(\sqrt{1-1/x}-1+1/(2x)\big){\rm d}x - \int_\lambda^\infty \big(\sqrt{1-1/x}-1+1/(2x)\big){\rm d}x .$$ Then you can go on with the asymptotic expansion in the same way (the integrand is $\sim c/x^2$, that gives you a term of size $c/\lambda$, etc...) It's a general fact that you won't usually get a nice, explicit expression for the "constant term" in that way.

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  • $\begingroup$ Can you explain why we do not get the constant term? Is there no other way to get the constant term? What if the constant term is the leading term? $\endgroup$ – user111187 Feb 26 '15 at 21:19
  • $\begingroup$ You do get the constant term, but it is usually not explicit : here, it is the integral $C = \int_1^\infty (\sqrt{1-1/x}-1+1/(2x)){\rm d}x$. It so happens that we can evaluate it to $-\ln(2)+1/2$. For your second question, consider $J_3(\lambda)$ above. The "constant term" is the leading term, it is the limit as $\lambda\to\infty$ (that's $C$). The procedure is then the same : you substract it, and analyze the remainder (that's the last line). $\endgroup$ – Sary Feb 26 '15 at 21:45

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