[PDF] Asymptotic Expansion of Integrals - University of Utah





Previous PDF Next PDF



1 Intégrales généralisées

Exercice 1. Montrer que l'intégrale de f : t ?? exp(?t) est convergente sur [0 +?[ et. ? +?. 0 exp(?t)dt = 1. Correction : Pour tout x > 0



Intégrales convergentes

9 mai 2012 ?t>A t?e?t ? e?t/2 . Or l'intégrale ? +?. 1 e?t/2 dt converge. En effet :.



Intégrale de Gauss

Intégrale de Gauss La fonction (t x) ??. ? 1. 0 e?(t2+1)x2 ... e?(tx)2 dt ce qui



Formules de Taylor. Applications. 1 Formule de Taylor avec reste

La formule de Taylor avec reste intégral `a l'ordre n s'écrit alors : exp(x)=1+ n. ? k=1 xk k! + xn+1 n! ? 1. 0(1 ? t)n exp(tx) dt.



Correction de linterrogation

?t/2 ? e?t/2. Or l'intégrale ?. ?. A e?t/2dt converge d'après le théorème d'intégrabilité des fonctions exponentielles. Comme ?t ? 0



Résumé sur les Intégrales Impropres & exercices supplémentaires

f(t)dt est divergente. Exemples. (a). On a. / x. 0 e?tdt = 1 ? e?x. Comme lim x?+? e?x = 0 l'intégrale. / +?. 0 e?tdt est convergente et vaut.



EQUATIONS DIFFERENTIELLES I Définition et notation

La solution générale de cette équation sur I est : y0 = k×e-A(t) où A(t) est une primitive de a(t) sur I et 



Python MP PC

TSI Oral



Développement asymptotique de lintégrale de sin(t)/t

t dt. Yves Coudene 16/10/03. L'intégrale ? N. 0 sin t t L'intégrale ? N. 0 e?tx sint dt se calcule explicitement `a l'aide des complexes :.



Etude de la fonction Gamma ?

e?ttx?1 et pour tout x ? R fx : R?+. ? R t. ?? f(x



Asymptotic Expansion of Integrals - University of Utah

Apr 16 2017 · exp t x? t dt: One can show that asymptotically the solution satis es ypxq c ? 3 3? 4xexp 3 x 2 2{3 as xÝÑ8: 1 Asymptotic Notation We begin by de ning asymptotic notations and asymptotic expansion These are useful in describing the limiting behaviour of a function when the argument gets closer to a particular complex number typically 0 or



List of Integrals Containing exp(x) - Math info

Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx=



Table of Basic Integrals Basic Forms

e t2dt (60) Z xex dx= (x 1)ex (61) Z xe axdx= x a 1 a2 e (62) Z x2ex dx= x2 2x+ 2 ex (63) Z x2eax dx= x2 a ax 2x a2 + 2 a3 e (64) Z x3ex dx= x3 3x2 + 6x 6 ex (65) Z



List of integrals of exponential functions - Informa?ní systém

List of integrals of exponential functions 3 ( is the modified Bessel function of the first kind) References • Wolfram Mathematica Online Integrator (http:/

What are the different types of integrals?

Integrals Containing sin Integrals Containing tan Integrals Continaing sec Integrals Continaing csc Integrals Containing cot Inverse Trigonometric Functions Hyperbolic Functions

How do you evaluate a definite integral?

Evaluate the definite integral using substitution: ?2 1 1 x3e4x ? 2dx. Integrating functions of the form f(x) = 1 x or f(x) = x ? 1 result in the absolute value of the natural log function, as shown in the following rule. The following formula can be used to evaluate integrals in which the power is ? 1 and the power rule does not work.

What are double integrals?

Double Integrals: Surface Area Triple Integrals Gradient of a Scalar Function Line Integral of a Vector Field Line Integral of a Scalar Field Green's Theorem Divergence of a Vector Field

How do you integrate an exponential function?

Exponential functions can be integrated using the following formulas. Find the antiderivative of the exponential function e ? x. Use substitution, setting u = ? x, and then du = ? 1dx. Multiply the du equation by ? 1, so you now have ? du = dx. Then, Find the antiderivative of the function using substitution: x2e ? 2x3.

Asymptotic Expansion of Integrals

Chee Han Tan

Last modied : April 16, 2017

Abstract

These notes are largely based on the last 3 weeks ofMath 6720: Applied Complex Variables and Asymptotic Methodscourse, taught by Christel Hohenegger in Spring 2017 and Alexander Balk in Spring 2016, at the University of Utah. Additional examples/remarks/results from other sources are added as I see t, purely for my own understanding. These notes are by no means ac- curate or applicable, and any mistakes here are of course my own. Please report any typographical errors or mathematical fallacy to me by emailtan@math.utah.edu

Motivation

A solid understanding in the asymptotic theory of integrals has proven to be invaluable for applied mathematician, but why integrals? The reason is that many functions that arise frequently in math- ematics, physics and engineering are dened by (complicated) integral expressions, and in most cases one resorts to numerical techniques to study these integrals due to the diculty in gauging its be- haviour. It is precisely this reason that many powerful analytical tools are developed to extract asymptotic behaviour of these integral functions for small or large values of the parameter. We brie y mention a few applications:

1.Integral transforms.The Fourier transform of a given functionfpxqis given by

fpq » R nfpxqei2xdx; and one is interested in estimating asymptotically ^fpqasÝÑ 8. For simple functions such as e |x|andex2, one can use contour integration to compute their Fourier transform explicitly. ForL1functions, Riemann-Lebesgue lemma states that^fpq ÝÑ0 as|| ÝÑ 8. However, these approaches are not available for complicated functions such as bump functions.

2.Special functions.Examples of special functions in mathematical physics include:

Airy function : Aipxq 1

8 0 cos xtt33 dt

Gamma function : pzq »

8 0 tz1etdt;Repzq ¡0: Airy functions appear in optics, electromagnetism, uid dynamics and nonlinear wave propaga- tion. 1

2 1 ASYMPTOTIC NOTATION

3.Dierential equations.In special cases, one might have an integral representation for solution

of ODEs and PDEs. Long time behaviour of the system can be understand using asymptotic expansion techniques. Consider the following initial value problem: #xy3pxq 2y0; yp0q 0; yp8q 0:

Its solution has an integral representation

ypxq » 8 0 exp tx?t dt: One can show that asymptotically the solution satises ypxq c 3

3?4xexp

3x2 2{3 asxÝÑ 8:

1 Asymptotic Notation

We begin by dening asymptotic notations and asymptotic expansion. These are useful in describing the limiting behaviour of a function when the argument gets closer to a particular complex number, typically 0 or8.

Denition 1.1.Letfpzq;pzqbe functions dened on

€C.

1. W esa ythat fpzq OppzqqaszÝÑz0if there exists a constantC¡0 and a neighbourhood

Uofz0such that

XU: i.e.fpzqpzq is bounded locally aroundz0. 2.

W esa yth atfpzq oppzqqaszÝÑz0if

lim zÑz0fpzqpzq0: 3. A sequence of gauge functions tnu8n0is anasymptotic sequenceaszÝÑz0if n1pzq opnpzqqaszÝÑz0for everyn0;1;:::: 4. The function fpzqis said to have anasymptotic representation (expansion) fpzq fNpzq N¸ n0a nnpzqaszÝÑz0; if for everyN0;1;2;:::, we have fpzq fNpzq opNpzqqaszÝÑz0: In other words, asymptotic representation of a function describes its asymptotic behaviour in terms of asymptotic sequence.

1 ASYMPTOTIC NOTATION 3

Remark 1.2.

1. In tuitively,an asymptotic e xpansionof a giv enfunction fis a nite sum which might diverges, yet it still provides an increasingly accurate description of the asymptotic behaviour off. There is a caveat here: for a divergent asymptotic expansion, for somez, there exists an optimal N

0N0pzqthat gives best approximation tof,i.e.adding more terms actually gives worse

accuracy. 2. Ho wever,for v aluesof zsuciently close to the limiting valuez0, the optimal number of terms required increases,i.e.for every"¡0, there exists anand an optimalN0N0pqsuch that fpzq N¸ k0a kkpzq "for every|zz0| andN¡N0: 3. One should think of fNpzqas converging for xedNin the limit aszÝÑz0. Observe that the denition of asymptotic expansion implies that the remainder term is \small" compared to the last termNpzqoffNpzq. Example 1.3.The functionskpxq xkform an asymptotic sequence asxÝÑ0and in this case the asymptotic representation is often called anasymptotic power series. The functions kpxq xkform an asymptotic sequence asxÝÑ 8. Proposition 1.4.Consider nding the leading asymptotic behaviour of the integral

Ipxq »

b a fpx;tqdtasxÝÑx0: Iffpx;tq f0ptqasxÝÑx0uniformly fortP ra;bs, i.e. lim xÑx0fpx;tq f0ptqf

0ptq0uniformly int;

then the leading behaviour ofIpxqasxÝÑx0is

Ipxq »

b a fpx;tqdt» b a f

0ptqdtasxÝÑx0;

provided that the integral on the RHS is nite and nonzero. Example 1.5.For instance, to determine the leading behaviour of the integral

Ipxq »

2 0 cosxt2x2t1{3dtasxÝÑ0; we simply setx0 and obtain

Ipxq »

2 0 cosp0qdt2 asxÝÑ0:

4 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS

2 Series Expansions and Integration By Parts

Broadly speaking, there are two ways of approximating a function: 1.

A con vergentexpansion, or

2.

A div ergentasymptotic expansion.

A convergent expansion can be easily obtained by integrating term by term the power series represen-

tation of the integrand, while a divergent expansion is usually constructed using integration by parts.

Depending on the limiting value, one is more favourable than the other.

Recall the Gamma function

pzq » 8 0 tz1etdt;Repzq ¡0: One can show using integration by parts that the Gamma function satises the functional equation zpzq pz1q; which can be used to uniquely extend pzqto a meromorphic function onC, with simple poles at the non-positive integersz:::;2;1;0. We note that: 12 8 0e t?t dt2» 0

8eu2du?:

For notational convenience, dene the following function: pz;xq » x 0 tz1etdt(Lower incomplete Gamma function) pz;xq » 8 x tz1etdt(Upper incomplete Gamma function)

Lemma 2.1.For anyn¥3,»8

0 etndtn1n wherepqis the Gamma function. Proof.We make a change of variableutn, thenduntn1dtnun1{ndt. The integral becomes: 8 0 etndt» 8 0 eudunu n1{n 1n 8 0 upn1q{neudt 1n 1n n1n Example 2.2.Consider approximating theerror function

Erfpxq 2?

x 0 et2dtasxÝÑ 8:

2 SERIES EXPANSIONS AND INTEGRATION BY PARTS 5

1.Convergent expansionThe Taylor series of the integrandet2aroundx0 is

e t21t2t42! t63! and it converges everywhere because it has no singularity as a function of complex variable. Integrating the Taylor series term by term, we obtain:

Erfpxq 2?

xx33 x510 x742 and this power series converges everywhere. However, the convergence is slow for large values ofxand it doesn't capture the asymptotic behaviour of ErfpxqasxÝÑ 8.

2.Divergent expansionWe rst rewrite the error function as follows:

Erfpxq 2?

»8 0 et2dt» 8 x et2dt 12? 8 x et2dt; Observe that the integrand is almost negligible ift¡¡x, so it contributes most to the integral whentis close tox. Using the identity e t2 12tddt et2 and integration by parts, we obtain for everyn¥0:

Gpnq:»

8 xe t2t ndt» 8 x 12tn1 ddt ret2s dt et22tn1 8 x 8 xpn1qet22tn2dt ex22xn1n12 Gpn2q 12 ex2x n1 pn1qGpn2q Thus, 8 x et2dtGp0q ex22x12 Gp2q ex22xex22p2x3q12 32
Gp4q ex22xex22

2x3p1qp3qex22

3x5p1qp3qp5q2

3Gp6q Repeating this procedure using the recurrence relation forGpnq, we nally obtain:

Erfpxq 12ex2?

12x14x3138x513516x7:::

12ex2?

N n0p1qnp2n1q!!2 n11x

2n1RN1pxq

6 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS

1ex2? N n0p1qnp2n1q!!2 n1x

2n1RN1pxq;

wherep2n1q!! p2n1qp2n3q:::p3qp1qandRNpxqis the remainder term, having the form R

N1pxq p1qN1?

p2N1q!!2

N1Gp2pN1qq:

Clearly, the series diverges for any largex. For the asymptotic sequence npxq ex2x

2n1asxÝÑ 8;

for a xedNwe have that |RN1pxq| C» 8 xe t2t 2N2dt C» 8 x

12t2N3

ddt ret2s dt

ÀCex2x

2N3

CNpxqx

2; whereCis a constant depending only onN. Hence, we show that

Erfpxq 1ex2?

N n0p1qnp2n1q!!2 n1x

2n1asxÝÑ 8:

To conclude, we present a table with number of terms required for both convergent and divergent expansion to approximate Erfpxqwithin 105, for dierent values ofx. It clearly suggests that the

divergent expansion is a better approximation compare to the convergent expansion.Convergent expansionDivergent expansion

Rangex 1x 2x 3x 5x¡3x¡2:5# of terms816317523 Example 2.3.Consider approximating the exponential integral: E

1pxq »

8 xe tt dtasxÝÑ 8:

Using integration by parts,

E

1pxq »

8 xe tt dt ett 8 x 8 xe tt

2dtexx

8 xe tt 2dt exx ett 2 8 x 2» 8 xe tt

3dtexx

exx

22»

8 xe tt 3dt

2 SERIES EXPANSIONS AND INTEGRATION BY PARTS 7

exN¸ n1p1qn1pn1q!x n p1qNN!» 8 xe tt

N1dtloooooooooooomoooooooooooon

R Npxq

For the asymptotic sequence

npxq exx n;asxÝÑ 8; We claim that the rst sum is an asymptotic expansion ofE1pxqasxÝÑ 8, with respect to the asymptotic sequencekpxq ex{xk. A coarse estimate on the remainder gives:

N1»

8 x etdtN!exx N1N!x Npxq:

Hence,

E

1pxq exN¸

k1p1qk1pk1q!k nN¸ k1a kkpxqasxÝÑ 8: Note that the asymptotic expansion diverges asNÝÑ 8for a xedx. We next approximateE1pxqasxÝÑ0. DierentiatingE1pxqusing Leibniz rule gives: dE 1dx exx 1x

1xx22!

x33!

Integrating term by term, we obtain

E

1pxq Clnxxx24

To ndC, we take the limit asxÝÑ0:

ClimxÑ0

»8 xe tt dtlnx

0:57772:

Note that

1p1q limzÑ0

z1z limnÑ8 lnnn¸ k11k

Example 2.4.Consider approximating the integral

Ipxq »

x 0 t1{2etdt 12 ;x asxÝÑ 8: The Taylor series of the integrandt1{2etaroundt0 is t

1{2et2t1{2t1{212

t3{216 t5{2:::; and it converges for allt0. Integrating this term by term gives: x 0 t1{2etdt2x1{223 x3{215 x5{2121 x7{2:::;

8 2 SERIES EXPANSIONS AND INTEGRATION BY PARTS

and it doesn't capture the asymptotic behaviour of 12 ;xasxÝÑ 8. On the other hand, a direct integration by parts gives: x 0 t1{2etdt t1{2etx 0 12 x 0 t3{2etdt; which diverges upon evaluating the boundary term att0. To nd an expansion that is useful for largex, we rewrite 12 ;xas follows: 12 ;x 12 12quotesdbs_dbs22.pdfusesText_28
[PDF] integrale sin(t)/t^2

[PDF] integrale sin(t)/t

[PDF] procédés théatraux

[PDF] tendinopathie genou traitement

[PDF] tendinite demi membraneux

[PDF] comment soigner une fabella

[PDF] fabella douloureuse

[PDF] tendinite poplité traitement

[PDF] mecanique de fluide resume

[PDF] mécanique des fluides bernoulli exercices corrigés

[PDF] fiche résumé mécanique des fluides

[PDF] mécanique des fluides cours pdf

[PDF] question ? choix multiple culture générale

[PDF] question ? choix multiple definition

[PDF] choix multiple orthographe