MATHEMATICS BONUS FILES for faculty and students http://www2onuedu/~mcaragiu1/bonus_fileshtml RECEIVED: November 1 2007 PUBLISHED: November 7 2007 Solving integrals by differentiation with respect to a parameter Khristo N Boyadzhiev Department of Mathematics Ohio Northern University Ada OH 45810 k-boyadzhiev@onuedu Contents 1 Introduction 2 Examples 3Using differential equations 4 Advanced technique using the Leibniz Integral Rule 5 Theory References
1 Introduction The 66th Annual William Lowell Putnam Mathematical Competition (2005) includes the evaluation of the following integral (A5) (11) (A solution is provided in [11]) We shall evaluate it now by the very natural and powerful method of Differentiation with respect to a parameter Define the function (12) Then (13) which is easily evaluated by partial fractions (14) Integrating we find (15) and setting we arrive at the equation from which 1
(16) This example illustrates the method very well Many integrals containing log arctan arcsin can be evaluated this way Note Equation (15) can be used to evaluate (17) where is Catalan s constant (see [1]) (18) In what follows we present several examples involving the differentiation on a parameter method Some theorems ensuring the legitimacy of the work are listed at the end Applying the theorems in every particular case is left to the reader Our main reference in the excellent book of Fikhtengolts [5] 2 Examples Example 21 Consider the integral (211) where Differentiating for we find: 2
and the substitution turns this integral into Therefore (212) In order to evaluate the constant we factor out on the left hand side and on the right hand side of (212) then split both sides this way canceling the term now and setting we find that As a result two integrals are evaluated (213) (214) In particular for : (215) Example 22 We evaluate here the convergent improper integral 3
(221) Differentiation for gives: where we use the substitution to obtain Therefore (222) and in particular for : (223) This integral can be found for instance in [6 p 608] It also appears on p321 in [3] with a suggestion to be evaluated by numerical experimentation The evaluation of as given here is from [4] Note that the similar integral (224) 4
can not be evaluated in the same way Its derivative is not easy to integrate Later we shall evaluate (224) by a more sophisticated method (see Section 5) An integral similar to (223) is (225) which however can not be evaluated like (223) The value of this integral is where is the Catalan constant - see (18) Example 23 Another convergent improper integral is (231) We find (232) ie (233) Therefore (234) 5
as both sides are zeros when When (235) Note that one needs for the evaluation in (232) but this restriction later drops and (234) hold for every Also the reader may wonder why we use instead of in (231) Repeating the above work with instead of will provide an answer to this question Example 24 We shall evaluate now two other integrals with arctangent The first one is (241) For all we compute Therefore (242) Comparing this to (234) we conclude that for all (243) Related to (241) is the following integral 6
(244) Differentiating for the first variable and integrating by parts for Setting here we find so finally (cf [8 p 506]) (245) Example 25 The next example is a very popular one (251) where is the parameter and is fixed The derivative uniformly convergent for is: Using integration by parts or a table one finds 7
and therefore (252) (the constant of integration is zero because ) At this point we can set and take limits of both sides We arrive at the classical result (253) true for all Note that we can use the derivative for in (251) and evaluate the integral in a similar manner In the variable the integral represents the Laplace transform of Example 26 A similar integral is (261) where again is fixed The derivative is and Setting we find and finally 8
(262) Example 27 A symmetrical version of Example 26 is the integral (271) defined for Note that the integral is divergent at infinity when although its derivative exists for We shall require Integrating this derivative we obtain (272) Comparing (272) to (262) we see that for all Example 28 Using the well-known Euler-Poisson integral (281) we can evaluate the following 9
where Computing the derivative we find: = and therefore (282) as both sides are zero for Example 29 Sometimes one can use partial derivatives as in the following example Consider the integral (291) with four parameters We shall use only the last two The partial derivatives are: It is easy now to restore the function from these derivatives where is unknown The integral (291) is zero when and therefore 10
Finally (292) Example 210 The last example in this section is a very interesting one It comes from the nice classical book of Woods [13 p143] Let (2101) Obviously We can assume that in the following computations The last integral can be evaluated by setting and then After some algebra (2102) From here we derive when and when Correspondingly Since we have 11
for all (2103) To determine we factor out inside the logarithm in (2101) split into two integrals and cancel the terms on both sides in the same manner as in Example 21 After that setting we obtain Therefore and since is continuous this extend also to ie (2104) 3 Using Differential equations Example 31 Sometimes differentiating with respect to a parameter can be combined with other methods like differential equations Here is a case in point Consider the integral (311) Here and integration by parts leads to the separable differential equation or with solution For we have (see (281)) Therefore 12
Example 32 Next we evaluate the Laplace integrals ( ) (321) which can be viewed as Fourier cosine and sine transforms of the functions and correspondingly We shall derive a second order differential equation for Obviously (322) but we can not differentiate this equation further as special trick Adding to both sides becomes divergent We apply instead a (323) (see (253)) we obtain and here we differentiate again for Thus we arrive at the second order differential equation with general solution ( - constants) Here as is a bounded function In order to find we set in (321) and evaluate Finally (323) and from (322) also 13
(324) These results can be used to evaluate some similar integrals For instance (325) by integrating (323) for and adjusting the constant of integration Differentiating this integral for we obtain also (326) Example 33 Similar to (321) are the following Laplace integrals and (331) which are also solved by a differential equation but not in the same manner Let We differentiate twice (332) and since at the same time we arrive at the differential equation (333) 14
This second order linear differential equation with constant coefficients can be solved by variation of parameters to obtain (334) where the solution involves the special sine and cosine integrals and The choice of integral limits here is dictated by the initial condition From equation (332) we find also (335) Example 34 We can evaluate Hecke s integral (341) by using a differential equation [7] Differentiating for (342) and substituting we obtain the equation ie (343) From here - a constant (344) 15
We can set now and using the fact that (345) find (346) 4 Advanced technique This method is based on the Leibniz Integral Rule: where are appropriate functions (see [12]) We shall evaluate now the integral (41) by using this rule Preliminary investigation shows that we can successfully work with the function where with Applying the theorem one finds 16
We focus now on the first term on the right hand side the integral Let us call it The substitution helps to solve it This function is easy to integrate as and hence one antiderivative is Also and we write Therefore integrating we obtain 17
and using the limit we evaluate Finally (42) In order to evaluate (1) we set here After some simple algebra we obtain (43) Remarks Integrating (1) by parts we arrive at therefore (44) With the substitution we obtain also the following result (see [4]) (45) 18
Problem Evaluate (46) by using the function (see [4]) Note that the somewhat similar integral (47) cannot be evaluated this way The value of this integral is ( see (18)) The substitution transforms (47) into which is the fourth in the list in [1] 5 Theorems 1 Suppose the function is defined and continuous on the rectangle together with its partial derivative In that case (51) In order to apply this theorem in the case of improper integrals we have to require uniform convergence of the integral with respect to A simple sufficient condition for uniform 19
convergence is the following 2 Suppose is continuous in on and is integrable on that interval If then the integral (52) is uniformly convergent on 3 Suppose the function is defined and continuous on the semi strip together with its partial derivative In that case when the first integral is convergent and the second is uniformly convergent for all The case of improper integrals on finite intervals is treated in a similar way For details and proofs we refer to [2] [5] [7] and [13] The book [5] presents the Leibniz Rule in full details References 1 Victor Adamchik Integral and Series Representations for Catalan's Constant http://wwwcscmuedu/~adamchik/articles/catalanhtm 2 T Apostol Mathematical Analysis Addison Wesley Publishing Co 2nd ed 1974 3 J Borwein D Bradley R Girgensohn Experimentation in Mathematics A K Peters 2004 4 Khristo Boyadzhiev Hans Kappus; Solution to problem E 3140 Amer Math Monthly 95 (1) (1988) 57-59 20
5 G M Fikhtengolts A Course of differential and integral calculus (Russian) Vol 2 Nauka Moscow 1066 6 I S Gradshteyn and I M Ryzhik Tables of Integrals Series and Products Academic Press 1980 7 Omar Hijab Introduction to Calculus and Classical Analysis Springer 1997 8 A P Prudnikov Yu A Brychkov O I Marichev Integrals and Series Vol1: Elementary Functions Gordon and Breach 1986 9 Joseph Wiener Differentiation with respect to a parameter College Mathematics Journal 32 No 3(2001) pp 180-184 10 Joseph Wiener Donald P Skow Differentiating indefinite integrals with respect to a parameter Missouri Journal of Mathematical Sciences Vol 3 no 2 (1991) 65-69 11 66th Annual William Lowell Putnam Mathematical Competition Math Magazine 79 (2006) 76-79 12 Eric W Weisstein "Leibniz Integral Rule" From MathWorld--A Wolfram Web Resource http://mathworldwolframcom/leibnizintegralrulehtml 13 Woods F S Advanced Calculus New Edition A Course Arranged with Special Reference to the Needs of Students of Applied Mathematics Ginn and Co Boston MA: Ginn Boston 1954 Updated November 2007 21