Solutions to Problem Set 6
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1 Solutons to Problem Set 6 Problem 6. (Resdue theory) a) Problem Boas. n ths problem we wll solve ths ntegral: x sn x x + 4x + 5 dx: To solve ths usng the resdue theorem, we study ths complex ntegral: z z + 4z + 5 ez dz = where s a sem-crcle n the upper half complex plane. When the radus of the sem-crcle goes to nnty, the ntegral of the functon along the arc wll go to zero, and the ntegral along the real axs wll become ths ntegral: z + 4z + 5 dz: f we use Euler's dentty e z = cos z + sn z, then take the magnary part of the ntegral, and substtute z by x, we end up wth our orgnal ntegral. f we had taken the real part, we would have ended up wth the correspondng cosne ntegral. Anyway, n the end, we have to nd the resdues of f (z) at the poles n the upper half plane, and multply the sum of those resdues wth to get the value of the contour ntegral. The magnary part of ths s then the value of our ntegral. Note that ths demands that the degree of the polynomal n the denomnator s greater than (not equal to) the degree of the polynomal n the numerator - f not, the ntegral along the arc would not go to zero. Ths whole recpe s summarzed on page 69 n Boas, where there s a parameter m (whch s one n ths case) that allows for more general ntegrals to be solved. Factorng f (z), we locate the poles of the functon: f (z) = z + 4z + 5 = (z + + )( + ) : There are rst order poles at z =, and the pole at + s n the upper half plane. We need the resdue at ths pole: R(f (z); z = + ) = = lm The contour ntegral s then lm (z + )f (z) z! + z! + z + + = ( + )e ( + )e = = ( )e : Our ntegral s the magnary part of ths: x sn x x + 4x + 5 dx = m :
2 = (cos( )e sn( )e ) = (cos + sn ) : e b) Problem n Boas. n ths problem, we wll solve ths ntegral: cos(x) (4x + 9) dx: Ths ntegral has a very smlar form to the ntegral n the prevous problem, so we go drectly to the recpe on page 69 n Boas. The rst thng we notce s that the ntegrals gven there are between and, but our ntegral s from to. Luckly, the ntegrand s an even functon (f (x) = f ( x)), and they have ths nce property that we have used plenty of tme before: cos(x) (4x + 9) dx = (4x + 9) cos(x)dx: Now we can use the recpe. We check the three ponts:. Snce the numerator s a constant, we can regard t as a polynomal of degree zero. The denomnator s dentely a polynomal, of degree 4.. The numerator s postve for all x R, and so t has no real zeros. 3. The degree of the denomnator s 4, whch s greater than the degree of the numerator (whch s zero). m = n ths case. Wth these three condtons met, t s straght forward to solve the ntegral: just construct the correspondng complex ntegrand f (z) = (4z + 9) ez ; nd the resdues at the poles n the upper half plane, sum them and multply by, and then take the real part (snce we have a cosne n our orgnal ntegrand) to get the nal answer. However, even f t s straght forward, ths may nvolve a lot of work, and a lot of traps to fall nto. So let us get to t. We start by makng an equaton of the denomnator beng zero, to locate the poles and factor the denomnator: (4z + 9) = ) 4z + 9 = ) z = 3 ) f (z) = e z ((z 3 )(z + 3 )) : We see that we have two second order poles, at z = 3. Only the pole at z = 3 s n the upper half plane, so we only need to nd the resdue at ths pole. Usng the method for a second order pole, descrbed near the bottom of page 685 n Boas, we nd the resdue: = R f (z); z = 3 = d dz (z + 3 ) e z (z 3 )ez 6(z + 3 )4 z = 3 z 3 f (z)! z = 3 = (3) 3 6(3) 4 e 3 = e 3 54 :
3 Multplyng ths wth, we get the contour ntegral: = 54 e 3 : Then we take the real part, and multply by the factor /, due to our orgnal ntegraton lmt: cos(x) (4x + 9) dx = Re c) Problem n Boas. Here, we wll study ths ntegral: x sn(x) x dx: = 54 e 3 : Ths ntegral may look suspcous at rst sght - we are ntegratng through two ponts (x = ) n whch the ntegrand s not dened. Now, takng the lmts when x goes towards or, the numerator also goes to zero, and the lmts are dened. By lettng the ntegrand havng these values for x =, there s no problem wth the ntegraton doman. We solve ths by the same recpe as n the two prevous problems, wth one addtonal pont to the resdue theorem: f there are poles on the contour we ntegrate along, we only add half of the resdues there. Gong to the complex functon and factorng the denomnator, we have ths contour ntegral: = ( z)( + z) dz; where s a sem-crcle n the upper half plane wth a radus that wll go to nnty, as usual. The two rst order poles of the ntegrand, at z =, are on the contour. We calculate the resdues: R(f (z); z = ) = lm (z )f (z) = z! ; R(f (z); z = ) = : We add the resdues, dvde by, and multply by (or just smply multply by ): = + = : Now, snce we are nterested n the sne ntegral, we take the magnary part: x sn(x) x dx = m d) Problem n Boas. Now we wll study ths ntegral: x 4 dx: = : Agan there are two ponts n the ntegraton doman (x = ) where the ntegrand s not dened. Ths tme, however, the numerator does not go to 3
4 zero when x goes to, and so we can not smply extend the ntegrand wth addtonal values at these ponts. But we can nd a prncple value. Wth the resdue theorem, ths s straght forward. We rst observe that the ntegrand s an even functon, and so we have x 4 dx = x 4 dx We transform t to a contour ntegral, and factor the denomnator: = (z )(z + )(z )(z + ) dz; where s yet another sem-crcle n the upper half plane wth nnte radus. There s a rst order pole at z = nsde our contour, and two rst order poles, at z =, on the contour. We easly compute the resdues: R(f (z); z = ) = 4 ; R(f (z); z = ) = 4 ; R(f (z); z = ) = 4 : Now, the value of the contour ntegral s tmes the sum of the resdues nsde the contour, plus tmes the sum of the resdues on the contour. The sum of the resdues on the contour s, so we get = 4 = : Now, our orgnal ntegral s related to ths contour ntegral by a factor of /, and we only get a prncpal value: P V x dx = 4 4 : e) Here, we wll study ths ntegral: x + dx: We rst multply wth x, the complex conjugate of the denomnator, n the numerator and denomnator. Then we splt the ntegral n two: (x ) (x + )(x ) dx = x x + dx x + dx: These two ntegrals can be solved straght forward wth the recpe at page 69 n Boas: we solve these contour ntegrals by the resdue theorem ( have already factored the denomnators): f (z)dz = z (z )(z + ) ez dz; f (z)dz = (z )(z + ) ez dz; where s the usual sem-crcle, then we take the real part of both, multply the second by, and then sum up. 4
5 Both ntegrands have poles at z =, where the pole at z = s nsde the contour. The resdues are R(f (z); z = ) = lm(z )f (z) = lm z! z! z + = e R(f (z); z = ) = lm(z )f (z) = lm z! z! We then get these two values for the contour ntegrals: e z z + = f (z)dz = e ; f (z)dz = e : e : Now we take the real parts (snce we are dealng wth the cosne ntegrals): x x + dx = Re x + dx = Re f (z)dz f (z)dz = = e : Then we multply the second term by and add up to get the nal answer: Problem 6. (Tensors) x + dx = e : a) Problem.4.4 n Boas. Fndng the nerta tensor s straght forward by usng equaton (4.5). Snce we are workng wth a contnuous mass dstrbuton, we wll use the ntegral form. The most mportant to note here when comparng to example 5 are the ntegraton lmts. We are studyng a quarter of a sphere. n sphercal coordnates, wth x = r sn cos ; y = r sn sn and z = r cos, we have r [; ]; [; ] and [; ]. The xx component then s = xx = (y + z )dm = (r x )dv V r ( sn cos )r sn drdd = 5 ; where the ntegral s straght forward solved by usng onlne servces such as Wolfram Alpha, or by lookng up n tables. The yy and zz components are found smlarly, and both actually have the same value as xx. Now for the o-dagonal elements, rst xy : xy = xydm = (r 4 sn 4 sn cos )r sn drdd = 5 = yx: The last equalty comes from symmetry n the ntegrand. Smlarly, we nd xz = zx = yz = zy =. As we see, these o-dagonal elements are not equal 5
6 to xy - ths s what the warnng s referrng to. Wrtten as a matrx, the nerta tensor looks lke ths: 5 A : The prncpal moments of nerta are the egenvalues of. We can nd them ether by computer (matlab has very smple functons to compute those), or by hand, by use of the characterstc polynomal. They are 5 ; 5 ; + 5 The prncpal axes are the egenvectors of. Agan, we can nd them by usng a computer, or by hand, by settng up the equatons wth the egenvalues: v = v. You wll be left wth some degrees of freedom here, so have chosen the easest egenvectors by (my) eye. f you use matlab, you wll get normalzed egenvectors. got these three vectors (wrtten n artesan coordnates): (; ; ); (; ; ); (; ; ) : b) Problem.4.6 n Boas. As n the prevous problem, calculatng the nerta tensor n ths case s straght forward usng equaton (4.5). Ths s a dscrete case, and so we use the dscrete versons of the equatons. These are just straght forward sums, where m = for the rst mass, and m = for the second mass. x ; y and z are the gven vector components: xx = X m (y + z ) = 9: Smlarly, we get yy = 9 and zz = 6. For the o-dagonal elements, we get xy = X m x y = 3 = yx ; where the last equalty s yet due to symmetry n the sum. For the other components, we get xz = zx = and yz = zy =. So the nerta tensor looks lke ths: The prncpal moments of nerta are the egenvalues of the nerta tensor: they are f6; 6; g, and the prncpal axes are the egenvectors, where got these: : A : f(; ; ); (; ; ); (; ; )g: 6
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