Evaluation of integrals by differentiation with respect to a parameter

December 8 Evaluation of integrals by differentiation with respect to a parameter Khristo N Boyadzhiev Department of Mathematics and Statistics, Ohio Northern University, Ada, OH 458, USA k-boyadzhiev@onuedu Abstract We review a special technique for evaluating challenging integrals by providing a number of eamples Many of our eamples prove integrals from the popular table of Gradshteyn and Ryzhik Keywords Improper integrals; integrals depending on a parameter; integral evaluation; Laplace transform; Laplace integrals Mathematics subject classification 6A4; 3E; 33E Introduction and three eamples There are various methods for evaluating integrals: substitution, integration by parts, partial fractions, using the residue theorem, or Cauchy s integral formula, etc A beautiful special technique is the differentiation with respect to a parameter inside the integral We review this technique here by providing numerous eamples, many of which prove entries from the popular handbook of Gradshteyn and Ryzhik [6] In our eamples we focus on the formal manipulation Several theorems justifying the legitimacy of the work are listed at the end of the paper Applying the theorems in every particular case is left to the reader

We hope integral lovers will appreciate this review and many will use it as a helpful reference The eamples and techniques are accessible to advanced calculus students and can be applied in various projects Many more integrals from [6] can be proved by using the same approach We also want to mention that for many of the presented eamples, solving the integral by differentiation with respect to a parameter is possibly the best solution Our main reference is the ecellent book of Fikhtengolts [5] which is the source of several eamples Some integrals solved by this technique can be found in [] and [] The method is presented in various publications, for instance, [4], [7], [9], [], [], and [3] Below we evaluate three very different integrals in order to demonstrate the wide scope of the method In section we present a collection of eighteen more or less typical cases In section 3 we show how differential equations can be involved very effectively In section 4 we demonstrate a more sophisticated technique, where the parameter appears also in the integral limits Eample We start with a very simple eample It is easy to show that the popular integral () sin d is convergent In order to evaluate this integral we introduce the function () λ sin F( λ) = e d, λ > and differentiate this function to get λ F ( λ) = e sin d = + λ (Laplace transform of the sine function) Integrating back we find F( λ) = arctan( λ) + C

Setting λ yields the equation = + C, ie C=, and therefore, (4) λ sin F( λ) = e d = arctan λ (cf entry 394 in [6]) Taking limits for λ we find (5) sin d = Eample The 66 Annual William Lowell Putnam Mathematical Competition (5) included the integral (A5) (6) ln( + ) d + with a solution published in [] This is entry 49(8) in [6] We shall give a different solution by introducing a parameter Consider the function (7) F( λ) = ln( + λ) d + defined for λ Differentiating this function we get F ( λ) = d ( + λ)( + ) This integral is easy to evaluate by splitting the integrand in partial fractions The result is ln( + λ) λ F ( λ) = + ln + + λ + λ 4+ λ 3

Integrating we find (8) F λ ln( + ) ln ( λ) = d + arctan λ+ ln ( + λ ) + 8, and setting λ = we arrive at the equation That is, F() = ln 4 (9) ln( + ) d = ln + 8 As we shall see later, many integrals containing logarithms and inverse trigonometric functions can be evaluated by this method It is good to notice that integrating (9) by parts we find ln( + ) arctan d = ln( + ) arctan d + + and therefore, we have also arctan d = ln + 8 Eample 3 This is Problem 997 from the Mathematics Magazine 89(3), 6, p 3 Evaluate () e d Solution We show that for every λ > 4

() λ e F( λ) d = λln 4 Indeed, differentiating this function with respect to λ (which is legitimate, as the integral is uniformly convergent on every interval < a< λ < b) we find λ λ λ e λ e e F ( λ) = e d = d = ln by using Frullani s formula for the last equality (see below) We conclude that F( λ ) is a linear function and since F () = we can write F( λ) = λln 4 With λ = we find F () = ln 4 Frullani s formula says that for appropriate functions f( ) we have () f ( a) f ( b) b d = [ f () f ( )] ln a General eamples Eample We start this section with a simple and popular eample Consider the integral () J ( ) = d ln for Note that the integrand is a continuous function on [,] when it is defined as zero at = and as (its limit value) at = Since yields d = ln, differentiation with respect to d J ( ) = d= + 5

and J( ) = ln( + ) + C Since J () = we find C = and finally, () d = ln ( + ) ln The similar integral (3) β d ln where β, can be reduced to () by writing β β = ( ) Thus we have (4) + d = ln ln + β β Another way to approach () is to use the substitution Frullani integral = e t which transforms it to the (5) e ( + ) t t e t dt see () Eample We evaluate here the improper integral (6) J ( λ ) = arctan λ d Differentiation yields J ( λ) = d ( + λ ) and with the substitution = cosθ this transforms into 6

/ / dθ J ( λ) = dθ = + λ cos θ ( + tan θ + λ ) cos θ / d tanθ tanθ arctan + λ + tan θ + λ + λ + λ = = = Therefore, (7) J ( λ) ln ( λ λ ) = + +, since J () = In particular, with λ =, (8) arctan d = ln(+ ) This integral is entry 453() in [6] Note that the similar integral (9) J ( λ ) = arctan λ d cannot be evaluated in the same manner The derivative here becomes / d cosθ + λ + λ J ( λ) = = dθ = ln + λ + λ cos θ λ + λ + λ λ ( ) which is not easy to integrate The integral (9) will be evaluated later in section 4 by a more sophisticated method Eample 3 Now consider () J ( λ) = ln( + λ ) + d with 7

J ( λ) = λ + λ + ( )( ) d λ λ = d = = λ + λ + λ λ + λ under the restriction λ This way () J ( λ) = ln( + λ) We needed λ for the evaluation of J ( λ), but this restriction can later be dropped For equation () we only need λ > In particular, for λ =, ln( + ) d = ln + The similar integral + ln( ) 3 d = (ln ) + 4 48 is evaluated by the same method in [3] In that article one can find also the evaluation ln( + ) d = ln G, + where G is Catalan s constant (see the remark at the end of Section 4) Eample 4 We shall evaluate here two more integrals with arctangents The first one is 4535(7) from [6], arctan λ () G( λ) = d ( + ) For all λ >, λ we compute 8

d λ G ( λ) = = ( + λ )( + ) λ + + λ d λ = = λ + λ ( ), and hence (dropping the restriction λ ) (3) G( λ) = ln( + λ) and for λ = arctan (4) ln d = ( + ) Comparing this to () we conclude that for all λ > (5) ln( + λ ) arctan λ d = d = ln( + λ) + ( + ) Eample 5 Related to () is the following integral arctan( λ )arctan( µ ) (6) G( λµ, ) = d for λµ>, Using the evaluation (3) we write arctan( µ ) λ Gλ ( λµ, ) = d = ln + ( + λ ) µ and integrating this logarithm by parts with respect to λ we find G( λ, µ ) = [( λ + µ ) ln( λ + µ ) λln λ ] + C( µ ) 9

Setting λ yields C( µ ) = µ ln µ Finally, arctan( )arctan( ) (7) λ µ [( ) ln( ) ln ln ] d = λ + µ λ + µ λ λ µ µ Eample 6 Consider the integral 3943 from [6] β cos λ F( λ) = e d, where β > is fied We have β F ( λ) = e sin λ d = λ λ + β and integrating back F( λ) = ln( λ + β ) + C( β) To compute C( β ) we set λ = and this gives C( β) ln = β Therefore, (8) cos λ d λ β = ln + e β Eample 7 A symmetrical analog to the previous eample is the integral e F( λ) = cos β d, λ defined for λ and β The integral is divergent at infinity when β = We have

λ F ( λ) = e cos β d = λ λ + β and integrating (9) λ e λ F( λ) = cos β d = ln + β so that for any λ, β > λ e β cos λ cos β d = e d Note that the integral 395(3) from [6] λ µ e e cos β d λ λ µ µ can be reduced to (9) by writing e e = ( e ) + ( e ) and splitting it in two integrals Thus λ µ e e µ + β cos β d = ln λ + β Eample 8 Using the well-known Gaussian integral, also known as the Euler-Poisson integral, () e d= we can evaluate for every λ the integral e F( λ) = d λ Indeed, we have for λ >

λ, λ λ F ( λ) = e d = ep( ( λ) ) d λ = so that () λ e d λ F( λ) = = Eample 9 Sometimes we can use partial derivatives as in the following integral Consider the function () p λ e cos q e cos µ F( λµ, ) = d with four parameters Here λ >, µ will be variables and p>, q will be fied The partial derivatives are λ λ, λ + µ F ( λµ, ) = e cos µ d = λ λ µ λ + µ F ( λµ, ) = e sin µ d = µ It is easy to restore the function from these derivatives F( λµ, ) = ln( λ + µ ) + C( pq, ), where C( pq, ) is unknown The integral () is zero when λ = p and µ = q, so from the last equation we find C( pq, ) = ln( p + q) / Therefore, (3) p λ e cos q e cos µ λ + µ d = ln p + q In all following eamples containing e λ we assume λ >

Eample Now consider λ sin( a)sin( b) J ( λ) = e d where a > b> are constants Clearly, λ J ( λ) = e sin( a)sin( b) d λ λ = e cos ( a b) d e cos ( a b) d + λ λ = λ + ( a+ b) λ + ( a b) Integrating with respect to λ and evaluating the constant of integration with λ we find (4) λ sin( a)sin( b) λ + ( a + b) 4 λ + ( a b) J ( λ) = e d = ln This is entry 3947() in [6] Eample Using the previous eample we can evaluate also entry 3947() in [6] λ sin( a) sin( b) G( λ) = e d where again a > b> We have from above λ + ( a+ b) λ + ( a b) G ( λ) = J( λ) = ln = ln 4 λ + ( a b) 4 λ + ( a+ b) and integrating by parts, 3

λ λ + ( a b) λ λ G( λ) = ln d λ 4 λ + ( a+ b) 4 λ + ( a b) λ + ( a+ b) With simple algebra we find λ λ ( a+ b) ( a b) = λ + ( a b) λ + ( a+ b) λ + ( a+ b) λ + ( a b) and the integration becomes easy The result is λ λ + ( a b) a b λ a+ b λ b 4 λ + ( a+ b) a b a+ b G( λ) = ln + arctan arctan + (the constant of integration is found by setting λ ) This answer is simpler than the one given in [6] With λ = we prove also 374(3) from [6] sin( a) sin( b) b d = ( a b > ) Eample Similar to (4) is the integral λ sin( a)cos( b) G( λ) = e d (this is 3947(3) in [6]) Suppose a > b> Then λ G ( λ) = e sin( a)cos( b) d λ λ = e sin ( a b) d e sin ( a b) d + + a+ b a b = + λ + ( a+ b) λ + ( a b), 4

and after integration with respect to λ, λ λ G( λ) = arctan + arctan a+ b a b, where the constant of integration / is found by letting λ Setting b a we find also λ sin( a)cos( a) λ e d = arctan 4 a Using the identity sin( a)cos( a) = sin( a) this integral can be reduced to (4) Eample 3 λ cos( a) cos( b) (5) F( λ) = e d (entry 3948(3) in [6]) Differentiating we find λ cos( b) cos( a) λ + a F ( λ) = e d = ln λ + b in view of (8), as cos( b) cos( a) = cos( b) + cos( a) Integration by parts yields (6) λ λ + a b a F( λ) = ln + barctan aarctan b λ + λ λ (again the constant of integration is found by setting λ ) Various integrals with similar structure can be evaluated by this method or by reducing to those already evaluated here For eample, entry 3948 (4) from [6] λ sin ( a) sin ( b) A( λ) = e d, can be reduced to (5) by using the identity sin cos = The results is 5

λ λ + 4b a b A( λ) = ln + aarctan barctan 4 λ + 4a λ λ Now we shall evaluate several integrals involving logarithms of trigonometric functions Eample 4 Consider the integral (7) ( ) = ln( cos ) J θ dθ for > Differentiation with respect to yields dθ cos θ J ( ) = and then the substitution = tanθ turns this into d J ( ) = = arctan = + Therefore, ( ) ln( cos θ) θ ln ( ) J = d = + + C In order to evaluate the constant of integration we write this equation in the form (factoring out in the left hand side and in the right hand side) cos θ ln + ln dθ ln ln C = + + + Removing ln from both sides and setting we compute C = ln As a result, two integrals are evaluated For the second one we set β = / in (7) 6

(8) + ln( cos θ) dθ = ln ( > ) (9) + β ln( cos ) d ln ( ) β θ θ = β In particular, with β = in (9) we obtain the important log-sine integral (3) ln(sin θ) dθ = ln Eample 5 In the same way we can prove that (3) / + θ θ = ln( sin ) d ln + + for any > Calling this integral F( ) and differentiating we find F ( ) = / sin θ dθ + sin θ Now we divide top and bottom of the integrand by cos θ and then use the substitution = tanθ F ( ) = d + + + ( )( ( ) ) Assuming for the moment that and using partial fraction we write F ( ) = + + ( + ) d + = arctan arctan( ) + = = + + + 7

Simple algebra shows that F ( ) = (+ + ) + which is eactly the derivative of ln(+ + ) Thus F( ) = ln(+ + ) + C Setting = we find C = ln and (3) is proved Eample 6 Let < Now we prove the interesting integral, entry 4397 (3) in [6] (3) ln(+ cos θ) F( ) dθ = arcsin cosθ Assuming that the value of the integrand at θ = / is, the integrand becomes a continuous function on [, ] Then F ( ) dθ + cosθ θ t dt which is easily solved with the substitution tan = t, so that cos θ =, dθ = + t + t Thus dt F ( ) = = + + ( ) t + dt + t + = arctan t =, + and (3) follows since both sides in this equation are zeros for = Eample 7 Let again < Using the results from the previous eample we can prove 8

(33) + ln( + cos θ) dθ = ln Indeed, let this integral be F( ) We have cosθ + cosθ dθ F ( ) = dθ = dθ = + cosθ + cosθ + cosθ = (for the moment ) Integration is easy: ( ) d F( ) = ln + = ln + ln( + ) + C = ln (+ ) + C Now we can drop the restriction With = we find C = ln and (33) is proved Eample 8 The last eample in this section is a very interesting integral It can be found, for eample, in the book [] on p 43 (34) F = + d ( ) ln( cos ) We first assume and Then cos + F ( ) = d = d cos+ cos+ d cos + = 9

The last integral can be evaluated by setting as before tan = t, with t dt cos =, d= + t + t Some simple work gives + F ( ) = arctan t and we find from here that F ( ) = when < and F ( ) = when > Thus F( ) = C when < and definition (3) shows) we have F( ) ln = C + for > Since F () = (as the F( ) = for all < Net, to determine the constant C when > we factor out inside the logarithm in (34) and write cos, F( ) = ln + d = ln + F = ln + = ln that is, C = and = for F( ) ln > This also etends to = ±, ie = for F( ) ln It is good to mention here that the evaluation of this integral for the case < can be done immediately by using the series representation for (35) k cos k + = k ln( cos ) k = The case > can be reduced to this one by writing ln( cos + ) = ln cos + = ln + ln cos +

The integral can be written in a symmetric form with β (cf entry 636(4) in [8]) (36) ln( β β cos + ) d = ln Eample 3 Consider the integral 3 Using differential equations t y( ) = e cos ( t) dt Here t =, y ( ) te sin ( t) dt and integration by parts leads to the separable differential equation dy y = y or = y d with general solution y( ) = Ce For = in the original integral we have y() = according to () Therefore, y( ) = e cos ( t) dt = e t With a simple rescaling of the variable we can write this result as a cos ( ) t /4 a e t d = e ( a > ) a

This last integral was used in the solution of Problem 896 of the Mathematics Magazine (vol 83, June 3, 8-3) Eample 3 In this eample we evaluate two interesting integrals (373, () and (3) in [6]) cos λ sin λ F( λ) = d, and G( λ) = d a + a +, which can be viewed as Fourier cosine and sine transforms We shall use a second order differential equation for F( λ ) First we have (3) F ( λ) = G( λ) We cannot differentiate further, because G ( λ) is divergent Instead, we shall use a special trick, adding to both sides of (3) the number sin = d (see (5)) After a simple calculation sin λ F ( λ) + = a d a ( + ) Differentiating again we come to the second order differential equation F = af with general solution aλ aλ F( λ) = Ae + Be, where AB, are arbitrary constants Suppose a > and λ Then A =, because F( λ ) is a bounded function when λ To find B we set λ = and use the fact that

d B= F() = = arctan = + a a a a Finally, cos λ (3) F( λ) = d = e a + a and from (3) we find also aλ sin λ aλ (33) G( λ) = d = e a + This result can be used to evaluate some similar integrals Integrating (3) with respect to λ and adjusting the constant of integration we find entry 375 () [6] sin λ aλ d = ( e ) a ( + ) a Differentiating this integral with respect to a we prove also entry 3735 sin λ ( a λ λ aλ d = e ) e 4 3 a ( + ) a 4a Eample 33 We shall evaluate two Laplace integrals For s > and a > consider st st e te F() s = dt,and G() s = dt a + t a + t Differentiating twice the first one we find (34) F () s = Gs (), F = G () s and at the same time 3

st st te a + a + t e () () + + ( ) G s = dt = dt = a F s + a t a t s which leads to the second order differential equation F + af= s This equation can be solved by variation of parameters The solution is F( s) = [ci( as)sin( as) si( as)cos( as)] a involving the special sine and cosine integrals sin t sin t si( ) = dt = + dt t t, ci( ) cost dt t = The choice of integral limits here is dictated by the initial conditions F( ) = G( ) = From (34) we find also G( s) = ci( as)cos( as) si( as)sin( as) The integral Fs () can be used to give an interesting etension of the integral (5) For any ab>, we compute sin( a) d ( ) sin( a t b t bt ) e + dt d e sin( a ) d e = = dt + b bt e dt ( ) ci( )sin( ) si( )cos( ) t + a = a = af b = ab ab ab ab This is entry 377 () from [6] Taking limits of both sides when b yields (5) In the same way we prove entry 37(3) 4

cos ( a) d = ci( ab )cos( ab ) si( ab )sin( ab ) + b Eample 34 We shall evaluate Hecke s integral d H ( ) = ep d, > by using a differential equation (cf [7]) Differentiation yields d H ( ) = ep d and then the substitution = / t leads to the separable differential equation dh d H ( ) = H( ), ie = H with solution H( ) = M ep( ), M > a constant Setting here = and using the fact that H () =Γ (/ ) = we find (35) H ( ) ep( ) = Eample 35 Consider the two integrals ([5], p73) (36) and U ( ) = ep( )cos d V ( ) = ep( )sin d We differentiate U ( ) and set y = / to find 5

U ( ) = ep( )sin d ep sin( y ) dy = y = ep sin( y ) dy y A second differentiation gives U ( ) = ep sin( y ) dy 4 ep( )sin d = y y, that is U ( ) = 4 V( ) In the same way we compute V ( ) = 4 U( ) We define now the comple function W ( ) = U ( ) + iv ( ) This function satisfies the second order differential equation W = 4iW with characteristic equation r + 4i = and roots r = + i and r = i From these roots we construct the general solution to the differential equation W( ) = Aep( r) + Bep( r) with parameters A and B Eplicitly, W( ) = Aep( )(cos + isin ) + B ep( )(cos isin ) At this point we conclude that B =, since W ( ) is a bounded function Setting = we find W() = A At the same time by the definition (36) of the above integrals W () U () ep( ) d = = = and 6

W( ) = ep( )(cos( ) + isin( )) From here, comparing real and imaginary parts we conclude that (37) U ( ) = ep( )cos( ), V ( ) = ep ( )sin( ) 4 Advanced techniques In certain cases we can use the Leibniz Integral Rule [5], [9]: d d ψ( ) ψ( ) d f (, ) d = f (, ) d + f ( ψ, ( )) ψ ( ) f ( ϕ, ( )) ϕ ( ) d, ϕ ( ) ϕ ( ) where f(, ), ϕ ( ), ψ ( ) are appropriate functions Eample 4 We shall evaluate the integral arctan d by using the function J( ) = ϕ ( ) arctan( ) d where > and ϕ ( ) = = with ϕ ( ) = Applying the Leibniz rule we find 7

arctan J ( ) = d ϕ ( )( + ) Let us call this integral A( ) We shall evaluate it by the substitution u, u = > : d ϕ ( ) ( + ) A( ) = = du + u + + u + + = ln = ln + + u + + This function is easy to integrate, as d + + ln = d + + and therefore, one antiderivative is 8 + + + A( ) d = ln We also have d d arctan = and therefore, Now we can integrate J ( ) to obtain ( ) arctan arctan d = 8

( ) + + J( ) = ln arctan + C 8 + Using the limit lim J ( ) = we find C = Finally, 8 arctan( ) + + (4) d ln ( arctan ) ϕ ( ) = + 8 + 8 Setting here we find after simple computation arctan (4) d = ( ln( + ) ) With = in (4) we find also 8 arctan( ) 5 + d = ln ( arctan 3) 8 + 5 8 3/ Remark Integrating by parts in (4) we find arctan arcsin d= d 8 + and therefore, Using the identity for >, we can write ( ) arcsin d = ln( + ) + arctan = arctan 9

arctan arctan d= d 4 In the second integral we make the substitution = / t to find also arctan t (43) dt = + ( ln( + ) ) 8 t t This integral was evaluated in [4] independently of (4) by using the same method Eample 4 We shall evaluate in eplicit form the function F( ) = ln( + ) / ( + ) d where > Differentiating by the Leibniz rule we compute F ( ) = d + / ( ) (notice that the function + becomes zero for = / ) ln( ) To solve the integral we first write it in the form d F ( ) = ( + ) / and then we make the substitution = t, t > to get dt + + F ( ) = = ln + t + + This function is easy to integrate, as 3

d + + ln = d + + Thus we find ( ) + + F( ) = ln = ln( + + ) ln( + ) + 8 8 (the constant of integration is zero, since F( ) for ) Simplifying this we get F = + + ( ) ln ( ) since + = = + + + + ln( ) ln ln( ) Therefore, for any >, (44) ln( + ) d = ln ( + + ) / ( + ) In particular, for =, (45) ln( + ) d = ln ( + ) ( + ) For = / in (44) we find (46) ln( + 4) 5 5 + d = ln ln + ln ( + ) Remark Note that the similar integral (47) ln( ) + d cannot be evaluated this way The value of this integral is G, where 3

( ) G = = + + n, n= (n + ) 3 5 is Catalan s constant The substitution = cosh t with ln( ) + = t turns (4,7) into t dt = G cosh t which is a well-known result 5 Some theorems Theorem A Suppose the function f(, ) is defined and continuous on the rectangle [, ab] [, cd] together with its partial derivative f (, ) Then d d f (, ) d = f (, ) d d c d c In order to apply this theorem in the case of improper integrals we have to require uniform convergence of the integral with respect to the variable A simple sufficient condition for uniform convergence is the following theorem Theorem B Suppose f(, ) is continuous on [ ab, ] [, ) and g ( ) is integrable on [, ) If f(, ) g ( ) for all a b and all, then the integral f (, ) d is uniformly convergent on [ ab, ] 3

Theorem C Suppose the function f(, ) is continuous on [ ab, ] [, ) together with its partial derivative f (, ) In this case d d f (, ) d = f (, ) d when the first integral is convergent and the second is uniformly convergent on [ ab, ] The case of improper integrals on finite intervals is treated in the same way For details and proofs we refer to [5], [7], [9], and [] The book [5] presents the Leibniz rule in full detail References [] Khristo N Boyadzhiev, Some integrals related to the Basel problem, SCIENTIA Series A: Mathematical Sciences, 6 (5), -3 Also arxiv:6357 [mathnt] [] Khristo N Boyadzhiev, On a series of Furdui and Qin and some related integrals,, arxiv:3468v3 [mathnt] [3] Khristo Boyadzhiev, Hans Kappus, Solution to problem E 34, Amer Math Monthly, 95 (), (988), 57-59 [4] Hongwey Chen, Parametric differentiation and integration, Internat J Math Ed Sci Tech 4 (4) (9), 559-579 [5] G M Fikhtengol ts, A Course of Differential and Integral Calculus (Russian), Vol, Nauka, Moscow, 966 Abbreviated version in English: The Fundamentals of Mathematical analysis, Vol, Pergamon Press, 965 [6] I S Gradshteyn and I M Ryzhik, Tables of Integrals, Series, and Products, Academic Press, 98 [7] Omar Hijab, Introduction to Calculus and Classical Analysis, Springer, 997 [8] A P Prudnikov, Yu A Brychkov, O I Marichev, Integrals and Series, vol Elementary Functions, Gordon and Breach 986 [9] Ioannis Markos Roussos, Improper Riemann Integrals, Chapman and Hall/CRC, 4 33

[] Joseph Wiener, Differentiation with respect to a parameter, College Mathematics Journal, 3, No 3(), pp 8-84 [] 66 th Annual William Lowell Putnam Mathematical Competition, Math Magazine, 79 (6), 76-79 [] Frederick S Woods, Advanced Calculus: A Course Arranged with Special Reference to the Needs of Students of Applied Mathematics New Edition, Ginn and Co, Boston, MA, 934 [3] Aurel J Zajta, Sudhir K Goel, Parametric integration techniques, Math Magazine, 6(5) (989), 38-3 34