Benjamin, Arthur T.

Dresden, Gregory P.

English

Keck Graduate Institute

Claremont Colleges
Scholarship @ Claremont
All HMC Faculty Publications and Research HMC Faculty Scholarship
4-1-2007
A Combinatorial Proof of Vandermonde's
Determinant
Arthur T. Benjamin
Harvey Mudd College
Gregory P. Dresden
Washington and Lee University
This Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusion
in All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contact
scholarship@cuc.claremont.edu.
Recommended Citation
Benjamin, Arthur T. and Gregory P. Dresden. "A Combinatorial Proof of Vandermonde's Determinant." The American Mathematical
Monthly, Vol. 114, No. 4, pp. 338-341, April 2007.
A Combinatorial Proof of Vandermonde's Determinant
Author(s): Arthur T. Benjamin and Gregory P. Dresden
Source: The American Mathematical Monthly, Vol. 114, No. 4 (Apr., 2007), pp. 338-341
Published by: Mathematical Association of America
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Accessed: 10/06/2013 18:56
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NOTES 
Edited by Ed Scheinerman 
A Combinatorial Proof of 
Vandermonde's Determinant 
Arthur T. Benjamin and Gregory P. Dresden 
We offer a combinatorial method of evaluating Vandermonde's determinant, 
1 Xq Xq 
1 X\ x2 
1 Xj Xr^ 
I Xn 
= 
n (*/- */)> 
0<i<j<n 
that is as easy as playing cards. Let Vn denote the Vandermonde matrix with (/, y)th 
entry vtj = x\ (0 
< i, j < n). Since the determinant of Vn is a polynomial in 
jco, x\, ... ,xn, it suffices to prove the identity for positive integers x0, x\, 
... ,xn 
with xo < x\ < < xn. We define a Vandermonde card to possess a suit and a value, 
where a card of Suit i has a value from the set {1, ... , x,}. (In our examples, we will 
let Suits 0, 1, 2, 3, and 4 be represented by suits O, A, O, ??, and 41, respectively.) 
Hence there are x0 + x{ -\-+ xn different Vandermonde cards, but we have at our 
disposal an unlimited supply of each. First we do some card counting. 
Card Counting Question 1. How many ways can Vandermonde cards be arranged in 
n-\-l rows, where row 0 is empty, row 1 has one card of Suit 1, row 2 has two cards of 
Suit 2, row 3 has three cards of Suit 3, ..., and row n has n cards of Suit nl The order 
of the cards is important, and we are allowed to repeat values of cards within each 
row. We call such an arrangement a Vandermonde table associated with the identity 
permutation n ? 012... n, an example of which is given in Figure 1. 
RowO 
Row 1 
Row 2 
Row 3 
Row 4 
Coll 
cn * 
C21 O 
C31<? 
C41 * 
Col 2 
C22 O 
c32V 
C42 
Col 3 Col 4 
C33V 
C43 * C44 4fc 
en e {1.xi) 
cij e {1.x2} 
C3j e {1.X3 
CAj G {1. x4} 
permutation n 
tt(0) =0-0 
tt(1) = 1 =* 
7T(2) =2 = 0 
7T(3) = 3 = C? 
tt(4) = 4 = 
Figure 1. A Vandermonde table associated with the identity permutation n = 01234 (or tt 
= OA<X?4*). Each 
2 3 4, 
X\X2X3XA of the i cards in row i has Suit i and a value from {1, ..., x?}. Such a table can be created in xxx\x\x\ ways. 
338 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 114 
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Answer. For / = 0, 1, ... , n, the / cards in row i all have Suit i, so their values can 
be assigned x\ ways. Hence, the number of arrangements is lxxx\x\ -x", which is 
the product of the diagonal entries of Vn. 
Card Counting Question 2. This is the same as Question 1, but now we are given 
a permutation n of the numbers 0 through n, say n 
= a0ax ...an. Here, row / must 
contain / cards from Suit n(i) ? ai. We call such an arrangement a Wandermonde table 
with permutation tt. A typical table is shown in Figure 2. 
.1 Y2 r3 Answer. Counting row by row again, there are lx7Jr(1)^7r(2)^7r(3) 
which is the product of the n + 1 entries of the form vn^)j from Vn. 
x^{n) such tables, 
RowO 
Coll Col 2 Col 3 Col 4 permutation tt 
ti(0) = 3 = <? 
Row 1 
Row 2 
Row 3 
Row 4 
cn * 
C21 O 
C31 O 
C41 A 
C22 O 
C32 0 
C42 * 
Cn G il. ..X4} 
C33 o 
C43 * C44 * 
c2j e {\...xo} 
C3j e {1.X2] 
CAj G {1.*ll 
7T(1)=4 = # 
71(2) =0 = 0 
7t(3) = 2 = 0 
tt(4) = 1 =* 
Figure 2. A Vandermonde table associated with permutation 7r = 34021 (or n 
= ??# O <>Jk). Each of the i 
2V3V4 , cards in row i has Suit n(i) and a value from {1, ..., xn(?)}. Such a table can be created in x4XqX^ ways, 
Card Counting Question 3. Same as Question 2, but now ix is not prescribed in 
advance, so tt can be any permutation of {0, 
... , n}. As before, each row is assigned 
a different suit and row / contains / cards of the assigned suit. For this unrestricted 
problem, such an arrangement is simply called a Vandermonde table. 
Answer. Sum the answer to Question 2 over all possible permutations of 0, ... ,n.ln 
other words, the number of ways to create a Vandermonde table is the permanent of 
Vn. 
Card Counting Question 4. Question 3 again, but now we count those arrangements 
with even permutations positively and those arrangements with odd permutations neg 
atively. 
Answer. By definition, this is the determinant of Vn. 
It remains to show that the answer to Question 4 also equals Y\o<i<j<n(xj 
~ 
xt). For 
a given Vandermonde table C let the cards of row / be denoted by CiX, Ci2, ... , Cn, 
with values ciX, ci2, ..., ciZ. We say that card Qy is small if ctj 
< 
Xj-\. For example, 
any card in column 1 with a value less than or equal to x0 (such as any card of Suit 0) 
is small. 
Card Counting Question 5. How many Vandermonde tables have no small cards? 
Answer. Let C be a Vandermonde table with no small cards. Since column 1 must not 
contain any cards of Suit 0, Suit 0 must be assigned to the empty row 0. Next, since 
column 2 must not contain any cards with value less than or equal to xx (such as any 
April 2007] NOTES 339 
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card of suit 1), we must assign suit 1 to row 1. Continuing this reasoning, row 2 must 
have Suit 2, ... , and row n must have Suit n. Thus C must be associated with the 
identity permutation. Furthermore, to avoid small cards in the first column, the values 
of the cards C\\, ..., Cn\ can be assigned in (jci 
? 
xo)(x2 
? 
*o)fe 
? 
*o) (xn 
? 
*o) 
ways. Likewise, the values of the cards in the second column can be assigned in (x2 
? 
X\)(x3 
? 
x\) 
' ' ' 
(*n 
? 
x\) ways, and so on down to the single card of Suit n in the last 
column, with a value that can be assigned in xn 
? xn_\ ways. We conclude that there 
are]!. 0<i<j<ny Xj 
? 
x?) Vandermonde tables with no small cards. 
We say that a Vandermonde table is good if it has no small cards and is bad if it has 
at least one small card. Note that since the identity permutation is even, all of the good 
tables are counted positively in the determinant of Vn. 
To complete the proof of Vandermonde's expansion, it suffices to show that every 
bad Vandermonde table can be paired up with another bad Vandermonde table with 
a permutation of opposite parity. Thus, when the determinant of Vn sums over all 
Vandermonde tables, the bad tables cancel each other out. When the dust settles, only 
the good tables (all counted positively) remain standing. 
Now let C be a bad Vandermonde table with permutation n = a?a\ ... an. We define 
Ihe first small card of C to be the small card ctj where j is as small as possible, and if 
column j has more than one small card, then we choose / to be as large as possible. In 
other words, we look for small cards from bottom to top, beginning in column 1. 
Coll Col 2 Col ? - 1 Colfc permutation n 
Row k - 1 
Row k 
Q-l,l * Ck-l,k-\6 
ck\V CklV Ckk <? 
*(*-!) = * 
n{k) = <? 
Figure 3. When the first small card occurs in the first column at card Q i, simply swap the cards of row k 
? 1 
with the cards C\a, , Ckk, and change the suit of card Ck\ 
Suppose that the first small card of C occurs in column 1, say card Ck\ for some 
1 < k < n. Then, since Ck\ is small, ck\ < x0, and since it is the first small card, there 
are no small cards below it; that is, when / > k, cn > x0. For definiteness, suppose 
that the cards in row k ? 1 have Suit n(k 
? 
1) = 41 and that the cards in row k have 
Suit 71 (k) = ??. (We make no assumptions about the suit number for hearts or spades.) 
Now consider the Vandermonde table C obtained by swapping all ^ 
? 1 spade cards 
with all of the heart cards except for card Ck\. Then change the suit of card Ck\ from 
hearts to spades. The suit change from hearts to spaces is legal because ck\ < x0, which 
is a legal value for all suits. (Here we are exploiting the fact that x0 < x\ < < xn.) 
Notice that Ck\ is still the first small card of C', albeit with a new suit, and thus if we 
apply the swapping procedure to C, we obtain C. That is, (C'Y = C. Furthermore, 
C has permutation n' 
= a0a\ ... akak_\ ... an. Permutations n and it' have opposite 
parity since they differ by the transposition of hearts and spades (see Figure 3). 
Now suppose that the first small card of C occurs in column j with j > 2, say 
at card Ckj. Then ckj < Jt/_i, and there are no small cards anywhere in columns 1 
through j 
? I nor below card Ckj in column j. As before, suppose that the cards of 
row k have the heart suit and that the cards of row k ? 1 have the spade suit. Create 
a new Vandermonde table C by swapping the first j 
? I cards of rows k ? 1 and k, 
340 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 114 
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Col 1 Col 7-I Col ; Col k 
- 1 Col k permutation ti 
Row k ? 1 | Cjt-i.i 
=3= 
Ck 
Row k ck\V 
X 
Cjt-1.7* Q-U-l* 
Q.;-l <? I Q/' <? I cu- <? 
7T(A: 
? 
1) = 
TT(k) = <? 
Figure 4. When Q/ is the first small card and j 
> 2, then swap the first j 
? 1 cards of row k ? 1 with the 
first j 
? 1 cards of row k, change the suit of card Ckj, then swap the remaining cards of rows k 
? 1 and k. In 
the new Vandermonde table, card Ckj remains the first small card. 
leaving card Ckj in its place, but changing its suit from hearts to spades, then swapping 
the remaining k 
? 
j cards of rows k 
? 1 and k, as in Figure 4. 
Why is it legal to change the suit of card Ckj from hearts to spades? Since Q7 was 
the first small card, then the spade card Ck-\j-\ is not small and therefore has a value 
strictly greater than X/_2- Thus all spade cards can take on values less than or equal to 
Xj-\. Since Ckj is small, its value is at most jc/-i, so changing it from hearts to spades 
is allowable. 
As before, Ckj remains the first small card of C, so (C)' 
? C and C has permuta 
tion tt', which has opposite parity of tt since they differ by a transposition. Thus there 
is a one-to-one correspondence between the positively counted Vandermonde tables 
with small cards and the negatively counted Vandermonde tables with small cards. 
Therefore the determinant of Vn is the number of Vandermonde tables with no small 
cards, namely, Y\o<i<j<n(xj 
~ 
xi)> as desired. 
Remark. For another combinatorial proof of Vandermonde's determinant, where the 
cancellation occurs in the product instead of the sums, see the short paper by Ira Gessel 
[1]. 
REFERENCES 
1. I. Gessel, Tournaments and Vandermonde's Determinant, J. Graph Theory 3 (1979) 305-307. 
Harvey Mudd College, Claremont, CA 91711 
benjamin @ math. hmc. edu 
Washington and Lee University, Lexington, VA 24450 
dresdeng @ wlu. edu 
Evaluation of Some Improper Integrals 
Involving Hyperbolic Functions 
Michael A. Allen 
In this note I present a result that seems elementary enough to be added to the list of 
tricks for evaluating integrals taught in a complex variables course, but one to which I 
have been unable to find any reference. It gives a straightforward procedure that can be 
used to evaluate a class of integrals some of which do not appear in [1] and for which 
Mathematica 5.1 [2] generates expressions involving exotic special functions that it 
cannot simplify further. 
April 2007] NOTES 341 
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A Combinatorial Proof of Vandermonde\u27s Determinant

Scholarship@Claremont

Claremont CollegesScholarship @ ClaremontAll HMC Faculty Publications and Research HMC Faculty Scholarship4-1-2007A Combinatorial Proof of Vandermonde'sDeterminantArthur T. BenjaminHarvey Mudd CollegeGregory P. DresdenWashington and Lee UniversityThis Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusionin All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contactscholarship@cuc.claremont.edu.Recommended CitationBenjamin, Arthur T. and Gregory P. Dresden. "A Combinatorial Proof of Vandermonde's Determinant." The American MathematicalMonthly, Vol. 114, No. 4, pp. 338-341, April 2007.A Combinatorial Proof of Vandermonde's DeterminantAuthor(s): Arthur T. Benjamin and Gregory P. DresdenSource: The American Mathematical Monthly, Vol. 114, No. 4 (Apr., 2007), pp. 338-341Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/27642196 .Accessed: 10/06/2013 18:56Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp .JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org. .Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toThe American Mathematical Monthly.http://www.jstor.org This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 18:56:04 PMAll use subject to JSTOR Terms and ConditionsNOTES Edited by Ed Scheinerman A Combinatorial Proof of Vandermonde's Determinant Arthur T. Benjamin and Gregory P. Dresden We offer a combinatorial method of evaluating Vandermonde's determinant, 1 Xq Xq 1 X\ x2 1 Xj Xr^ I Xn = n (*/- */)> 0<i<j<n that is as easy as playing cards. Let Vn denote the Vandermonde matrix with (/, y)th entry vtj = x\ (0 < i, j < n). Since the determinant of Vn is a polynomial in jco, x\, ... ,xn, it suffices to prove the identity for positive integers x0, x\, ... ,xn with xo < x\ < < xn. We define a Vandermonde card to possess a suit and a value, where a card of Suit i has a value from the set {1, ... , x,}. (In our examples, we will let Suits 0, 1, 2, 3, and 4 be represented by suits O, A, O, ??, and 41, respectively.) Hence there are x0 + x{ -\-+ xn different Vandermonde cards, but we have at our disposal an unlimited supply of each. First we do some card counting. Card Counting Question 1. How many ways can Vandermonde cards be arranged in n-\-l rows, where row 0 is empty, row 1 has one card of Suit 1, row 2 has two cards of Suit 2, row 3 has three cards of Suit 3, ..., and row n has n cards of Suit nl The order of the cards is important, and we are allowed to repeat values of cards within each row. We call such an arrangement a Vandermonde table associated with the identity permutation n ? 012... n, an example of which is given in Figure 1. RowO Row 1 Row 2 Row 3 Row 4 Coll cn * C21 O C31<? C41 * Col 2 C22 O c32V C42 Col 3 Col 4 C33V C43 * C44 4fc en e {1.xi) cij e {1.x2} C3j e {1.X3 CAj G {1. x4} permutation n tt(0) =0-0 tt(1) = 1 =* 7T(2) =2 = 0 7T(3) = 3 = C? tt(4) = 4 = Figure 1. A Vandermonde table associated with the identity permutation n = 01234 (or tt = OA<X?4*). Each 2 3 4, X\X2X3XA of the i cards in row i has Suit i and a value from {1, ..., x?}. Such a table can be created in xxx\x\x\ ways. 338 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 114 This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 18:56:04 PMAll use subject to JSTOR Terms and ConditionsAnswer. For / = 0, 1, ... , n, the / cards in row i all have Suit i, so their values can be assigned x\ ways. Hence, the number of arrangements is lxxx\x\ -x", which is the product of the diagonal entries of Vn. Card Counting Question 2. This is the same as Question 1, but now we are given a permutation n of the numbers 0 through n, say n = a0ax ...an. Here, row / must contain / cards from Suit n(i) ? ai. We call such an arrangement a Wandermonde table with permutation tt. A typical table is shown in Figure 2. .1 Y2 r3 Answer. Counting row by row again, there are lx7Jr(1)^7r(2)^7r(3) which is the product of the n + 1 entries of the form vn^)j from Vn. x^{n) such tables, RowO Coll Col 2 Col 3 Col 4 permutation tt ti(0) = 3 = <? Row 1 Row 2 Row 3 Row 4 cn * C21 O C31 O C41 A C22 O C32 0 C42 * Cn G il. ..X4} C33 o C43 * C44 * c2j e {\...xo} C3j e {1.X2] CAj G {1.*ll 7T(1)=4 = # 71(2) =0 = 0 7t(3) = 2 = 0 tt(4) = 1 =* Figure 2. A Vandermonde table associated with permutation 7r = 34021 (or n = ??# O <>Jk). Each of the i 2V3V4 , cards in row i has Suit n(i) and a value from {1, ..., xn(?)}. Such a table can be created in x4XqX^ ways, Card Counting Question 3. Same as Question 2, but now ix is not prescribed in advance, so tt can be any permutation of {0, ... , n}. As before, each row is assigned a different suit and row / contains / cards of the assigned suit. For this unrestricted problem, such an arrangement is simply called a Vandermonde table. Answer. Sum the answer to Question 2 over all possible permutations of 0, ... ,n.ln other words, the number of ways to create a Vandermonde table is the permanent of Vn. Card Counting Question 4. Question 3 again, but now we count those arrangements with even permutations positively and those arrangements with odd permutations neg atively. Answer. By definition, this is the determinant of Vn. It remains to show that the answer to Question 4 also equals Y\o<i<j<n(xj ~ xt). For a given Vandermonde table C let the cards of row / be denoted by CiX, Ci2, ... , Cn, with values ciX, ci2, ..., ciZ. We say that card Qy is small if ctj < Xj-\. For example, any card in column 1 with a value less than or equal to x0 (such as any card of Suit 0) is small. Card Counting Question 5. How many Vandermonde tables have no small cards? Answer. Let C be a Vandermonde table with no small cards. Since column 1 must not contain any cards of Suit 0, Suit 0 must be assigned to the empty row 0. Next, since column 2 must not contain any cards with value less than or equal to xx (such as any April 2007] NOTES 339 This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 18:56:04 PMAll use subject to JSTOR Terms and Conditionscard of suit 1), we must assign suit 1 to row 1. Continuing this reasoning, row 2 must have Suit 2, ... , and row n must have Suit n. Thus C must be associated with the identity permutation. Furthermore, to avoid small cards in the first column, the values of the cards C\\, ..., Cn\ can be assigned in (jci ? xo)(x2 ? *o)fe ? *o) (xn ? *o) ways. Likewise, the values of the cards in the second column can be assigned in (x2 ? X\)(x3 ? x\) ' ' ' (*n ? x\) ways, and so on down to the single card of Suit n in the last column, with a value that can be assigned in xn ? xn_\ ways. We conclude that there are]!. 0<i<j<ny Xj ? x?) Vandermonde tables with no small cards. We say that a Vandermonde table is good if it has no small cards and is bad if it has at least one small card. Note that since the identity permutation is even, all of the good tables are counted positively in the determinant of Vn. To complete the proof of Vandermonde's expansion, it suffices to show that every bad Vandermonde table can be paired up with another bad Vandermonde table with a permutation of opposite parity. Thus, when the determinant of Vn sums over all Vandermonde tables, the bad tables cancel each other out. When the dust settles, only the good tables (all counted positively) remain standing. Now let C be a bad Vandermonde table with permutation n = a?a\ ... an. We define Ihe first small card of C to be the small card ctj where j is as small as possible, and if column j has more than one small card, then we choose / to be as large as possible. In other words, we look for small cards from bottom to top, beginning in column 1. Coll Col 2 Col ? - 1 Colfc permutation n Row k - 1 Row k Q-l,l * Ck-l,k-\6 ck\V CklV Ckk <? *(*-!) = * n{k) = <? Figure 3. When the first small card occurs in the first column at card Q i, simply swap the cards of row k ? 1 with the cards C\a, , Ckk, and change the suit of card Ck\ Suppose that the first small card of C occurs in column 1, say card Ck\ for some 1 < k < n. Then, since Ck\ is small, ck\ < x0, and since it is the first small card, there are no small cards below it; that is, when / > k, cn > x0. For definiteness, suppose that the cards in row k ? 1 have Suit n(k ? 1) = 41 and that the cards in row k have Suit 71 (k) = ??. (We make no assumptions about the suit number for hearts or spades.) Now consider the Vandermonde table C obtained by swapping all ^ ? 1 spade cards with all of the heart cards except for card Ck\. Then change the suit of card Ck\ from hearts to spades. The suit change from hearts to spaces is legal because ck\ < x0, which is a legal value for all suits. (Here we are exploiting the fact that x0 < x\ < < xn.) Notice that Ck\ is still the first small card of C', albeit with a new suit, and thus if we apply the swapping procedure to C, we obtain C. That is, (C'Y = C. Furthermore, C has permutation n' = a0a\ ... akak_\ ... an. Permutations n and it' have opposite parity since they differ by the transposition of hearts and spades (see Figure 3). Now suppose that the first small card of C occurs in column j with j > 2, say at card Ckj. Then ckj < Jt/_i, and there are no small cards anywhere in columns 1 through j ? I nor below card Ckj in column j. As before, suppose that the cards of row k have the heart suit and that the cards of row k ? 1 have the spade suit. Create a new Vandermonde table C by swapping the first j ? I cards of rows k ? 1 and k, 340 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 114 This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 18:56:04 PMAll use subject to JSTOR Terms and ConditionsCol 1 Col 7-I Col ; Col k - 1 Col k permutation ti Row k ? 1 | Cjt-i.i =3= Ck Row k ck\V X Cjt-1.7* Q-U-l* Q.;-l <? I Q/' <? I cu- <? 7T(A: ? 1) = TT(k) = <? Figure 4. When Q/ is the first small card and j > 2, then swap the first j ? 1 cards of row k ? 1 with the first j ? 1 cards of row k, change the suit of card Ckj, then swap the remaining cards of rows k ? 1 and k. In the new Vandermonde table, card Ckj remains the first small card. leaving card Ckj in its place, but changing its suit from hearts to spades, then swapping the remaining k ? j cards of rows k ? 1 and k, as in Figure 4. Why is it legal to change the suit of card Ckj from hearts to spades? Since Q7 was the first small card, then the spade card Ck-\j-\ is not small and therefore has a value strictly greater than X/_2- Thus all spade cards can take on values less than or equal to Xj-\. Since Ckj is small, its value is at most jc/-i, so changing it from hearts to spades is allowable. As before, Ckj remains the first small card of C, so (C)' ? C and C has permuta tion tt', which has opposite parity of tt since they differ by a transposition. Thus there is a one-to-one correspondence between the positively counted Vandermonde tables with small cards and the negatively counted Vandermonde tables with small cards. Therefore the determinant of Vn is the number of Vandermonde tables with no small cards, namely, Y\o<i<j<n(xj ~ xi)> as desired. Remark. For another combinatorial proof of Vandermonde's determinant, where the cancellation occurs in the product instead of the sums, see the short paper by Ira Gessel [1]. REFERENCES 1. I. Gessel, Tournaments and Vandermonde's Determinant, J. Graph Theory 3 (1979) 305-307. Harvey Mudd College, Claremont, CA 91711 benjamin @ math. hmc. edu Washington and Lee University, Lexington, VA 24450 dresdeng @ wlu. edu Evaluation of Some Improper Integrals Involving Hyperbolic Functions Michael A. Allen In this note I present a result that seems elementary enough to be added to the list of tricks for evaluating integrals taught in a complex variables course, but one to which I have been unable to find any reference. It gives a straightforward procedure that can be used to evaluate a class of integrals some of which do not appear in [1] and for which Mathematica 5.1 [2] generates expressions involving exotic special functions that it cannot simplify further. April 2007] NOTES 341 This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 18:56:04 PMAll use subject to JSTOR Terms and Conditions

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A Combinatorial Proof of Vandermonde\u27s Determinant

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