The Tabular Method for Repeated Integration by Parts

1 The Tabular Method for Repeated Integration by PartsR. C. DailedaFebruary 21, 20181 Integration by PartsGiven two functionsf,gdefined on an open intervalI, letf=f(0),f(1),f(2),..,f(n)denotethe firstnderivatives off1andg=g(0),g( 1),g( 2),..,g( n)denotenantiderivatives main result is the following generalization of the standard Integration by Parts N, f(x)g(x)dx=n 1 j=0( 1)jf(j)(x)g( (j+1))(x) + ( 1)n f(n)(x)g( n)(x)dx.(1) induct onn. Whenn= 1 the formula becomes f(x)g(x)dx=f(x)g( 1)(x) f(1)(x)g( 1)(x)dxwhich is the result of Integration by Parts with the choicesu=fanddv=g now assume the formula (1) holds for somen 1. In the integral we integrate byparts, takingu=f(n)anddv=g( n)dx. Thendu=f(n+1)dxandv=g( (n+1))so that f(n)(x)g( n)(x)dx=f(n)(x)g( (n+1))(x) f(n+1)(x)g( (n+1))(x) this into (1) and collecting the integrated term into the sum we end up withn j=0( 1)jf(j)(x)g( (j+1))(x) + ( 1)n+1 f(n+1)(x)g( (n+1))(x)dx,which is precisely (1) withn+ 1 replacingn.

2 Hence then+ 1 case holds if thencase induction, the formula is valid for alln an immediate consequence we have the next result, which tells us how to obtain theantiderivative offgcompletely in the case thatfeventually differentiates to zero, whenfis a assumefis sufficiently smooth to be differentiated arbitrarily often this we need only assume thatgis continuous in mind that for a functionf, an exponent in parentheses indicates a derivative of a certain order, with negativeexponents indicating antiderivatives. We arenotsimply raisingfto thatf(n) 0. Then f(x)g(x)dx=n 1 j=0( 1)jf(j)(x)g( (j+1))(x) + result of Theorem 1 is perhaps most easily implemented using a table. In one columnwe listfand its firstnderivatives. In an adjacent column we listgand its firstnantideriva-tives. We label the columns asuanddvin keeping with the standard notation used whenintegrating by Parts .

3 We then multiply diagonally down and to the right to construct thesummands of (1), and then alternately add and subtract them to get the correct signs. Atthe final level, we multiply directly across, continue the alternation of signs, but integratethe resulting term to get the integral appearing in (1). See the diagram +%%gf(1) %%g( 1)f(2)+%%g( 2)f(3) $$g( 3)..f(n 1)( 1)n 1%%g( (n 1))f(n)( 1)n//g( n)Each solid arrow indicates a multiplication between the terms at either end, the result ofwhich is then added or subtracted to the to the antiderivative according to the sign shownwith the arrow. The final dashed arrow indicates that the two terms at either end are tobe multiplied together,integrated, and then added or subtracted from the antiderivativeaccording to the sign of ( 1)n. This gives usfg( 1) f(1)g( 2)+f(2)g( 3) f(3)g( 4)+ + ( 1)n 1f(n 1)g( n)+ ( 1)n f(n)g( n)dx,in agreement with (1).

4 Example (x3+ 2x 1) cos(4x). takef(x) =x3+ 2x 1,g(x) = cos(4x) and construct the table above:2udvx3+ 2x 1+''cos(4x)3x2+ 2 ''14sin(4x)6x+'' 116cos(4x)6 '' 164sin(4x)0+//1256cos(4x)The antiderivative is therefore14(x3+ 2x 1) sin(4x) +116(3x2+ 2) cos(4x) 3x32sin(4x) 3128cos(4x) + 0dx=(14x3+1332x 14)sin(4x) +(316x2+13128)cos(4x) +C. Example e2xsin(3x) takef(x) = sin(3x) andg(x) =e2xand construct the table of derivativesand antiderivatives:udvsin(3x)+%%e2x3 cos(3x) %%12e2x 9 sin(3x)+//14e2xThis tells us that e2xsin(3x)dx=12e2xsin(3x) 34e2xcos(3x) 94 e2xsin(3x) other words134 e2xsin(3x)dx=12e2xsin(3x) 34e2xcos(3x) +C3so that e2xsin(3x)dx=113e2x(2 sin(3x) 3 cos(3x)) +C. Example the 2-periodic cosine expansion of the functionx2(1 x), 0< x < half-range Fourier coefficients are given bya0=22 1 10x2(1 x)dx=x33 x44 10=112and forn 1an=21 10x2(1 x) cos(n x) compute the integral here using the Tabular technique, withf(x) =x2 x3andg(x) =cos(n x):udvx2 x3+''cos(n x)2x 3x2 ''1n sin(n x)2 6x+'' 1n2 2cos(n x) 6 '' 1n3 3sin(n x)0+//1n4 4cos(n x)Hencean= 2((x2 x3)sin(n x)n + (2x 3x2)cos(n x)n2 2 (2 6x)sin(n x)n3 3+ 6cos(n x)n4 4) 10= 2( cos(n )n2 2+ 6cos(n )n4 4 61n4 4)= 2( 1)n+1n2 2+ 12( 1)n 1n4 the cosine series is112+ n=1(2( 1)n+1n2 2+ 12( 1)n 1n4 4)cos(n x).

5 4 The plots below show the 10th, 20th, 30th and 40th partial sums of this series on the interval0 x 1, which more closely resemblef(x) =x2(1 x) as the number of terms increases,as they should. Example are numerous situations where Repeated Integration by Parts is calledfor, but in which the Tabular approach must be applied repeatedly. For example, considerthe integral (logx) we attempt Tabular Integration by Parts withf(x) = (logx)2andg(x) = 1 we obtainudv(logx)2+ 12 logxx //x5so that the antiderivative isx(logx)2 2 s no point in continuing the table, for if we do so we find that where ever we decide toterminate our columns we will be faced with the antiderivative logxdx. Evaluating thisrequires Integration by Parts . With the choicesf(x) = logxandg(x) = 1 we obtain thetableudvlogx+ 11x //xso that logxdx=xlogx dx=xlogx x+ this with our previous piece we find that (logx)2dx=x(logx)2 2xlogx+ 2x+C.

6 2 Application: Integrals of the Form P(x)T( x)dxwherePis aPolynomial andTis Sine or CosineConsider an integral of the form P(x)T( x)dxwherePis a polynomial andTis either sineor cosine. SinceP(n) 0 for every sufficiently largen, the corollary to Theorem 1 applies, the integral term in Theorem 1 reduces to a constant. Moreover, since the derivatives ofPare themselves polynomials while the antiderivatives of sine (or cosine) cycle through thefunctions sine, cosine and their negatives, if we integrate by Parts takingf(x) =P(x) andg(x) =T( x), we find that the antiderivative has the form4A(x) cos( x) +B(x) sin( x) +C(2)whereAandBare polynomials. There are two observations that can be made 1:AandBcan be computed algebraically, should one wish to avoid integra-tion by Parts . First, if we take the derivative of (2) we obtain (A (x) + B(x)) cos( x) +( A(x) +B (x)) sin( x).

7 Now suppose for the sake of argument thatTis we must haveA + B=Pand A+B = 0. Differentiating the first equation,multiplying the second by and adding we obtainA + 2A=P . Assuming is real,the characteristic equation of the complementary homogeneous ODEA + 2A= 0 hasonly complex roots. Hence, if we assumeAis a polynomial with degA= degP , we are4 This is not the only way to prove to find a unique solution toA + 2A=P by the Method of undeterminedcoefficients. OnceAis in hand,Bcan be found from the relationshipB= 1(P A ).It s worth noting, however, that Integration by Parts is probably far more efficient thanthe procedure we ve just 2:Once we ve computed P(x) sin( x)dxeither through Integration by partsor by using the procedure of the preceding observation, the antiderivative P(x) cos( x)dxcan be obtained from the antiderivative P(x) sin( x)dxby simply differentiating everyappearance of sine and cosine (formally), by replacing sine with cosine and cosine withnegative see this, suppose that P(x) sin( x)dx=A(x) cos( x) +B(x) sin( x) +C.

8 (3)Perform the formal differentiation of the trigonometric functions to obtain A(x) sin( x) +B(x) cos( x) +C.(4)Now derive:( A (x) B(x)) sin( x) + ( A(x) +B (x)) cos( x).This doesn t tell us much until we derive equation (3), too:P(x) sin( x) = ( A(x) +B (x)) sin( x) + (A (x) + B(x)) cos( x).Comparison of both sides shows thatP(x) = A(x) +B (x) andA (x) + B(x) = 0. Hencethe derivative of (4) is exactlyP(x) cos( x), as claimed. We summarize these findings withthe following diagram. P(x) sin( x)dxd =A(x) cos( x)d +B(x) sin( x)d P(x) cos( x)dx= A(x) sin( x)+B(x) cos( x)7