Solution to Problem 82-3: A definite integral

Solution to Problem 82-3: A definite integral

Citation for published version (APA):

Lossers, O. P. (1983). Solution to Problem 82-3: A definite integral. SIAM Review, 25(1), 101-102. https://doi.org/10.1137/1025013

DOI:

10.1137/1025013

Document status and date: Published: 01/01/1983

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PROBLEMS AND SOLUTIONS 101

zurAlgebraund Geometrie,toappear,containsthefollowingstatements: THEOREM 1.

f,(C)

2"- 1.

THEOREM2. 2

<--f2(K) <--

12.

Alsosolved by the proposer.

ADefiniteIntegral Problem 82-3,by A. J.

JERRI (Clarkson

College).

Prove

that

J(a(y,x)) J,(b(y,z))

J,(b(z,x))

Y

(y, x) (y,

z)

dy (z,

x)

0<b< a,

where

_J

(a(y,

x))/(y, x)

isthe x-"generalized"

_Jo-Hankel

typetranslationof

_J1

(ay)/y,

J

(a(y,

x))

ayJ (ay)

Jo(ax)

axJ

(ax)

Jo(ay)

(y,

s)

y

x

Solution by O. P. LOSSERS (Eindhoven University of Technology, Eindhoven, the

Netherlands).

From

Watson

[1,form.

5.11(8)]

it isreadilyseenthat

J

(a(y,

x))

fo

SJo(yS)jo(xs)ds.

(y,x)

Thustherequired integral, denotedby I,canbeexpressedas

I--

fo

y

dy

fO

SJo(yS)Jo(xs)ds

fo

tJo(yt)Jo(zt)dt

fo

SJo(xs)ds

foYJo(ys)dy

fo

tJo(yt)Jo(zt)dt.

Heretheinnerdouble integralisevaluatedbymeansof theinversiontheorem for Hankel transforms 1, form.

14.3(2)],

IJo(zs),

0<s<b,

fo

yJo(ys)dy

fo

tJo(yt)Jo(zt)dt

[0,

s>b.

Since 0<b<a, wethenfind

Jl(b(z,x)) I

]’bSJo(xS)Jo(zs)

ds

.to

(z,x)

REFERENCE

G. N.WATSON,ATreatiseontheTheoryofBesselFunctions,Cambridge UniversityPress,Cambridge,

1958.

Solutionbythe proposer.

Thisproblemisprovedby utilizingtheParseval equation of the

_Jo-Hankel

transform and recognizing that

J(ay)/y

and its "generalized"

_Jo-Hankel

translationJ(a(y,

x))/

(y,x) arethe

_J0-Hankel

transformsofp,(t) andJo(xt)p(t) respectively, wherepa(t) is

the gate function,

1,

pa(t)

102 PROBLEMSAND SOLUTIONS

Thesamemethodcanbe used for proving theHardyintegral,

f?

sin(y-x)a(y

x)

sin(y-z)b(y

z)

dy rsin

(z-x)

b(z

x)

O<b<a,

wherethe Parseval equation for theFouriertransformisemployed. Itis obviousthatthis

methodcanbe utilizedfor proving moregeneral integrals for the Hankel and for other transforms.

Inasecond solution,O. P. LOSSERS showsthat the stated equation holdsgenerally

for complexxandz.

AVolume Problem

Problem 82-4,by M. K.LEwis(Memorial University ofNewfoundland).

Anasymmetrically positioned hole ofradiusbisdrilled atrightanglestotheaxisofa solidrightcircularcylinder ofradius a

(a

>b).Ifthe distancebetween theaxisof thedrill

and theaxisof the cylinderisp,determinethevolumeofmaterialdrilled out.

Solution by J. BOERSMA, P. J. DE DOELDER and J. K. M. JANSEN (Department of

Mathematics and Computing Science, Eindhoven University ofTechnology,

Eind-hoven, theNetherlands).

Introduce Cartesian coordinates x,y,z, then the cylinder and the drill may be

describedby

(1)

Co:x

+ y2

a

2,

Cb:(X

p)2 +

z b

.

The volume ofthe intersection of

_Co

and

_Cb

is denoted by V(a,b,p). Without loss of generality,wemayassume0_-<b_-<a,p>_- 0.

Fromacross-sectionof

_Co

and

Cb

withtheplanez 0,it isreadilyseenthat

(2)

V(a, b, p) 2

f

_ib

(x

p)]l/

dxdy,

wherethedomainof integrationis

(3)

{(x,y)lx

+

y2

<=

a,lx

pl

<=

hi.

Wenowdistinguish threecases:

I. 0-_<_p_-< a b.Inthis case theintegral

(2)

reducesto

(4)

V(a,b,p) 4

[’p+b

[(a

x)(b

+

p

x)(x

+

b p)(x

+

a)]

/_dx.

Thelatter integralcan beexpressed in termsof elliptic integrals bymeans ofByrd and

Friedman [1, form.

254.38].

Omitting thedetailsofthe tediousthough straightforward calculation,wepresenttheresult

V(a,

b,p)

4(a

+

b

₊

p)-/

(a

+

b-

p)-/

(5)

[1/6 (p-

2ap 2a

+

2b-)(a

+

b

+

p)(p

+

a b)

K(k)

+ 1/6

(p +

2a

₊

2bZ)(a +

b

₊

p)(a

+

b p)E(k)

+

(a

-

b)p(p

₊

a-

b)II(a

2,

k)],

where

4ab 2b

(6)

k 0/2