Asymptotic behaviour/upper bound for $\int_0^{\infty} exp(-c x^a+K x^b)dx$ for $a>b>0$ as $K\rightarrow$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:49:55Z http://mathoverflow.net/feeds/question/104552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104552/asymptotic-behaviour-upper-bound-for-int-0-infty-exp-c-xak-xbdx-for-a Asymptotic behaviour/upper bound for $\int_0^{\infty} exp(-c x^a+K x^b)dx$ for $a>b>0$ as $K\rightarrow$ warsaga 2012-08-12T12:31:12Z 2012-08-13T17:33:37Z <p>What is theAsymptotic behaviour or an upper bound for $\int_0^{\infty} exp(-c x^a+K x^b)dx$ for $a>b>0$ as $K\rightarrow \infty$? Or any good reference for tools to tackle this question?</p> <p>I think the growth in $K$ should be polynomial because $-c x^a+K x^b=0$ yields $x=(K/C)^\frac{1}{a-b}$ on the range $[0,(K/C)^\frac{1}{a-b}]$ the maximum value of the integrand is again a power of K (take derivative and set 0) the product yields and upper bound on $\int_0^{(K/C)^\frac{1}{a-b}} exp(-c x^a+K x^b)dx$. On the other hand $\int_{(K/C)^\frac{1}{a-b}}^{\infty} exp(-c x^a+K x^b)dx$ should be decreasing in $K$ for large K.</p> <p>Thank you,</p> http://mathoverflow.net/questions/104552/asymptotic-behaviour-upper-bound-for-int-0-infty-exp-c-xak-xbdx-for-a/104561#104561 Answer by Alexandre Eremenko for Asymptotic behaviour/upper bound for $\int_0^{\infty} exp(-c x^a+K x^b)dx$ for $a>b>0$ as $K\rightarrow$ Alexandre Eremenko 2012-08-12T13:57:55Z 2012-08-12T13:57:55Z <p>This is a simple example for the Laplace Method of asymptotic evaluation of integrals. The essence of the method is that the main contribution to the integral comes from a small neighborhood of the critical point of the function under the exponent. Laplace method is explained in every serious calculus book. For more comprehensive treatment see the books of Fedoryuk, for example, Fedoryuk, M. V. (1987), Asymptotic: Integrals and Series, or here: <a href="http://www.encyclopediaofmath.org/index.php?title=Saddle_point_method" rel="nofollow">http://www.encyclopediaofmath.org/index.php?title=Saddle_point_method</a></p> http://mathoverflow.net/questions/104552/asymptotic-behaviour-upper-bound-for-int-0-infty-exp-c-xak-xbdx-for-a/104587#104587 Answer by Igor Rivin for Asymptotic behaviour/upper bound for $\int_0^{\infty} exp(-c x^a+K x^b)dx$ for $a>b>0$ as $K\rightarrow$ Igor Rivin 2012-08-13T00:26:06Z 2012-08-13T17:33:37Z <p>One can get a full asymptotic expansion as an application of <a href="http://en.wikipedia.org/wiki/Watson%27s_lemma" rel="nofollow">Watson's Lemma</a>. One need only observe that the integrand is maximized at $x_0 = \left(\frac{Kb}{ac}\right)^{1/(a-b)}.$</p> <p>Substituting $x = x_0 u,$ one gets(where $I$ is the original integral) $I = x_0 \int_0^\infty \exp(-c^{b/(a-b)} K^{a/(a-b)}((b/a)^{a/(a-b)} u^a - (b/a)^{b/(a-b)} u^b)) d u.$ Letting $t = c^{b/(a-b)} K^{a/(a-b)},$ the integral breaks up into two Watson Lemma integrals, one from $0$ to $1,$ the second from $1$ to $\infty.$ I leave the final computation of the asymptotics to the interested reader.</p> <p><strong>EDIT</strong> Actually, this is not quite Watson's lemma. You approximate the function $\phi(u) = (b/a)^{a/(a-b)} u^a - (b/a)^{b/(a-b)} u^b)$ by its Taylor series (at the maximum point $1).$ Since it is the maximum, this will look like $\phi(1) - (u-1)^2 \phi^{\prime\prime}(1)/2.$ This means that the integral is asymptotically approximated by a Gaussian integral (I am too lazy to compute $\phi^{\prime\prime}(1)$...\$)</p>