tT~

ON THE EXISTENCE OF THE MIXED DIFFERENTIAL COEFFICIENTS OF A REPEATED INTEGRAL.

By 6 a n e s h P r a s a d (Calcutta).

Adunanza del Io aprile zga7.

The object of the present paper is to apply PAVE DU BOIS-REYMO~qD'S Infinit~r- calcfil to decide whether the mixed differential coefficients of a repeated integral exist at a point where the integrand has a discontinuity of the second kind.

Although the question, whether the integral of a function of one variable can be differentiable at a point where the integrand has a discontinuity of the second kind, was first considered by Dilqi z) and TnOMAE 2) about fifty years ago and has been recently considered by L. NARAYAN a) and myself, the case of two or more variables has not been dealt with successfully by any previous writer. The results of the pre- sent paper are, therefore, believed to be new. For the sake of simplicity and fixity of ideas, I restrict the investigation to the case of two variables, take the repeated integral to be

fo' J?, F(x, y) ~ du t, u)dt,

the notion of integration being that of RmMA~N, and assume the point of disconti- nuity to be (o, o). Throughout this paper, the mixed differential coefficients at (o, o)

02F(o, o) and c3'F(° ' o) . Also +(t, u) as well as Z(t, u) is used are denoted by OxOy Oy~x

throughout to denote a function which is monotone in the neighbourhood of (o, o) and which tends to become infinite whether 4) t tends to o or u tends to o.

x) Fondamenti per la teorica delle funzioni di variabili reali, pp. 271-272. a) Einleitung in die Theorie der bestimmten Integrale, p. 17. a) Integration Images and the failure of the curvature formula (Proceedings of the Benares Mathe-

matical Society, vols. 4 and 5, x924 and I925). 4) The case in which the integrand has a discontinuity of the second kind, only when both t

and u tend to o, reduces to the case considered in my paper referred to in ~ 3 ; for this reason, it

has not been considered in the present paper.

x76 6 A N E S ~ VRASAD.

In § i, I generalise a criterion given by THOMAE S) for the case of one variable and prove that the generalised criterion furnishes a sufficient, but not necessary, con- dition for the existence of the mixed differential coefficients. In ~ 2, the case, in which f(t, u) is cos + (t, u), is considered. ~ 3 contains a treatment of the case in which f(t, u ) : z( t , u)cos + (t, u). The paper concludes with a detailed consideration of the cases in which the generalisaticm of THOMAE'S criterion fails.

General i sat ion of THOMAE'S criterion.

I. If

f xf(t~-u)dt, foYf(~U)du and t

exist, then c~'F(° ' o) OxOy exists and equals zero.

Proof: Let J(t, u) denote a function such that

0 __f(t, u) ot l(t' u ) - - t

Ydu fl~f(t, U)dt u do t

(.4)

and let K(t, u) denote a function such that

8 0 u K(t' u) __ - -

Then

Therefore

](t, . )

f" f(t, u)dt = xJ(x, u ) - l(t, u)dt (by integration by parts).

(i)

fo' fx F(x, y) ~ du f(t, u)dt

= d . xJ(x, . ) - - ](t, . )dr l

=xl.,(x, fo'-( x (by integration by parts).

5) Ueber die Differen~irbarkeit eines Integrales nacb der oberen Gren~e [Nachrichten d. KOniglichen Gesellschaft der Wissenschaften zu GSttingen, 0893), pp. 696-7oo ].

ON THE EXISTENCE OF THE MIXED DIFFERENTIAL COEFFICIENTS, ETC. 17~

Now, at (o, o),

at (x, o),

Thus

0 F __ 0F(o, o ) _ _ Lim F(o , y) - - F (o , o) Oy - - c)y y=o y

= o ;

0 F_ =-- 0 F(x, o) = Lira F(x, y) -- F(x, o) = Lim F(x, y) Oy Oy s=o y y=o y

0 F (x, o) O F (o, o)

02F(o, o) __ Lim Oy Oy Oxdy ,=o x

- - L i r a x ILim F (X' Y)I x=o X ~ y~o y

Therefore, from 0 ) ,

02F(o, o) I I Y ~'] - - - -Lim--rxK(x, o ) - - x L i m f - - f f K(x, .)dut--Limll f 'du f ](t, u)dt . OxOy ,=o x L ,=o ( y Jo } y=o t y .do do U

But, obviously, from the hypothesis in (A), J and K are continuous functions at (o, o). Therefore the first limit in the above expression inside the square brackets is equal to K(x, o) and the second is equal to

Therefore fo al(t, o)dt.

O'F(o, o)__ OxOy Lim.=o --xI foxf(t, o)dt=--J(o, o)--o.

Similarly 02 F(o , o) can be proved to be zero. OyOx 2. The condition in the criterion given above is not necessary for

of the mixed differential coefficients.

Proof: Consider the function f(t , u) given by

K(t, u)--cos[llog(~-~ + ufi~-)l*].

the existence

Then, obviously, the condition of the generalised criterion is not satisfied; for, K does not tend to any limit as t or u tends to zero and, consequently,

Now, fo Yclu f ' f ( t , U)d t u j o t d o e s n o t e x i s t .

$ ' K u't' I cos4 sin4 ~ t f(t, , , )=t =t"o--~-ff- ( . '+t') ' t +, + 7 " + ,

~aed. &re. Matcm Ps'~*r~o~ t. LII (19J8 ) . - Stamplto I1 ao marzo xga$, | j

z78 G A N E S H P R A S A 0 .

where + stands for !

llog (-~-~ + u-~)! 2.

Therefore f(t, u) is continuous in t, u at (o, o), and therefore the mixed differential coefficients of F(x, y) exist at (% o) and are zero.

S 2.

C a s e : / ( t , u) = cos + (t, u).

as

a+ t~7

(i)

8. The mixed differential coefficients

c9 + tends to infinity whether t tends to o or u tends to o.

Both the mixed differential coefficients are non-existent at (% o), provided

0+ tends to o whether t tends to o or u tends to zero. as well as u~-~

Proof of (B):

fo fo 2 cos + d t = 0-~--" cos ~ ,/t

o~t

= o ~ + | - - s i n + o ~ t i a + |

(by integration by parts)

sin+ / x sin t~ ~ t ( _ ~ _ ~ d t - - Oq,

exist at (o, o), provided that ~-F as well

(/3)

that

(¢)

I because, by hypothesis, ~ tends to zero as t tends to o.

Ot I

Also, by hypothesis, ~ tends to o as u tends to o ~lnd, consequently, its

Ot differential coefficient with respect to t tends to o as u tends to o. Therefore both the parts of the expression in ( 0 are continuous in . at u - - o and both tend to o

ON THE EXISTEI'CCE OF THE MIXED DIFFERENTIAL COEFFICIENTS, ETC. I79

with u; consequently

£ *cos 40, u)dt

is continuous in u at u - - o and tends to o with u. Therefore

OF(x, o) ~ - - - 0 .

Oy But

F ( o , o) O .

0y

Therefore c~2 F(o , o) is zero. ¢)x Oy

c~ = F(o , o) . Similarly, it can be proved that is zero.

OyOx

/ ' ; l( EXAMVLE. - - d u cos log ~ - og d

coefficients at (o, o) and they equal o.

4. Proof of (C) : Here, integrating by parts,

t has both the mixed differential

~' d u c o s F(x, y) = 4(t, ~)dt

fo' l I (2) -- du xcos 4 -}-- j . sin 4 d r

Y 0 4 . Y * 0 4 f d,, f t=-sin t. - - xycost~ Jr- x f u - - s m 4 d u . n t- 4 d Jo au Jo Jo ot

o+ o4 Now, by hypothesis, t~/- as well as u ~ tends to o whether t tends to o or u

0~F(o, o) depends on xy cos4, each of the tends to o. Therefore the existence of OxOy

c) ~ F(o , o) other two parts of F(x, y) in (2) contributing zero to the value of

Oxe)y O~ F(o, o)

Therefore is non-existent. OxOy '

Similarly, it can be proved that 0~F(o, o) is non-existent. e~yOx

EXAMPLE.- Both the mixed differential coefficients of

f f* I ( s d u cos log log -IT- + log d t It~'O ~ O

are non-existent at (% o).

t8o GANESH PRASAD.

§ 3 .

C a s e : f ( t , u) = Z (t, u) cos + (t, u).

If ~ as well as ~ tends 5. v T , . , y

03t d u

then both the mixed differential coefficients exist at (% o).

to o whether t tends to o or u tends to o 7

I f ( O --~+ "q t as t tends to o and -~+ -< u as u tends

03t 03u

( D )

to o, and (2) both

Z and ~ remain finite for every value of t and u, then both the mixed diffe- o+ 0 t 03u

rential coefficients exist at (% o), even if ~ does not tend to zero as u

03t

o and ~ does not tend to o as t tends to o. " - ' T

03u If ( i ) positive proper fractions k and 1 exist such that

and

and (2)

Z --~ 1- 03+ 03t

and (3)

and (4)

03t \ /

03+ 03u

uk >-- >-- i as u tends to .o

03t

- - ~ ~ i as t tends to % t z

c3u

as t tends to o and Z 03+ 03u

- - ~ i as u tends to o,

Z 03+ 03+ --¢ u 03t 03u

U

as u tends to o and --< t as t tends to %

as u tends to o and

03 z

o+ 03t

t

tends to

(E)

as t t e n d s to o 7

ON THE EXISTENCE OF THE MIXED DIFFERENTIAL COEFFICIENTS~ ETC. I8r

then both the mixed differential coefficients exist at (o, o).

(i)

Proof of (D) :

~ x

j z (t, u) cos + (t, . ) d t =

- - \ ~ 7 / (by integration by parts)

_ . Z (~ ,+ ~ _ u ) sin + - -

c~x f x ~ Oy (-~+ ~sin + dr,

to o. Also, by h y p o t h e s i s , -

(i~)

because, by hypothesis, ~ f f tends to o as t tends 0Z l

c3t o3t tends to o as u tends to o, and, consequently, its differential coefficient with respect to t tends to o as u tends to o. Therefore both the parts of the expression in ( I ) are continuous in u at u--" o and both tend to o with u; consequently

f xz (t, ~) cos + (t, . ) a t

is continuous in u at u - - o . Therefore

Therefore

c~ F (x , o ) = o. But a y

0F(o , o) __ o. &y

c) ' F (o , o)

does not affect the va-

O~F(o, o) is found to Oydx

- - - O . OxOy It will be noticed that changing the order of integration

lue of the repeated integral. Therefore, proceeding as above,

be equal to o. EXAMVLt.- The mixed differential coefficients of

exist at (% o) and equal o.

6. Proof of (E) : Starting from the equation ( I ) which we get as in the case of (D), and noting

that, from the hypothesis, it follows (i) that

I + ~ l o g t2 as t tends to o

i82 GANESH PRASAD.

and

and (ii) that we have

I + v-log ~ as u tends to %

f o x (t, u) cos ~b (t, u) d t - - * , (x, . ) sin d? (x, U) + Z (x, .) + (x, u) cos

c) F(x, o) is found to where * and ~ , are both finite for every value of x and u, c)y

be o. But 8 F ( o , o) 0"F(o , o) - - o. Hence - - o. Oy OxOy

Again, changing the order of integration does not affect the value of the repeated

integral. Therefore, proceeding as in the case of c~" F(o, o) 0 2 F(o, o) , we find that - - o . Oxe3y OyOx

EXA~tVLe.- The mixed differential coefficients of

exist at (o, o) and equal o.

7. Proof of (F ) : Starting from the equation (x) which we get as in the case of (D), we have

where

and

Z, standing for

Ydu Z - - - - F(x, y ) ~ (t, u )c os+ ( t , u)dt F, F~, o ' o g] o

fo' F, = Z, (x, . ) sin + (x, . ) d .

fo'£ F 2 --- d u Z2 (t, . ) sin '~ (t, u) d t,

O+(x, u) and Z2 for

0x \ ~ 7 /

Now, as regards F , , since Z, r - I and Z. ~ u, it follows, from a paper 6)

8 u

of mine to be published soon in CrELLE'S Journal, that 8 F , ( x , o) exists and is o. 8y

6) On the differentiability of the integral function.

ON THE EXISTENCE OF THE MIXED DIFFERENTIAL COEFFICIENTS~ ETC.

As regards F , , from the hypothesis, it follows that changing the order of inte- gration does not affect the value of F 2 . Thus

f f' xdt z (t, F2 = u) sin t~ (t, u)d u.

Denoting the integral with respect to u by ¢ ( t , y), we know that, since Z~ ~ I and

Z~ exists and is zero. Therefore 03 + q u, for every value of t, 03 go (t, o) 0y

o3u OF=(x, o3_ FtOo(t,

~y do : -~y °) tdt - o.

Thus 0*F(° ' o) exists and equals zero. OxOy

c~F(o, o) exists and equals zero. Similarly, it can be proved that O),e~x

EXAMVLE.- The mixed differential coefficients of

d u c o s t

I 0

exist at (o, o) and equal o.

F a i l u r e of t h e g e n e r a l i s a t i o n o f THOMAE'$ cr i te r ion .

provided that

and

8. The generalisation of TnOM*E's criterion fails, if

K(t , u) = cos + (t, u),

If

I + ~ log t~ as t tends to o

I log-~- as u tends to o. (G)

K(t , u) - - z( t , u) cos q, (t, u),

the generalisation fails, provided that

I I ( I ) Z -< l°g't2' as t tends to and -q l o g ~ as u tends to o,

I (2) + log -V

c3~+ I (3) X ~-F~u "< t-;

I as t tends to o and -< l o g ~ as u tends to o~

whether t tends to o or u tends to %

184 GANESH PRASAD.

and

(4) oCa+

Z~t Ou Proof of (G): Since

I tu

f(t, u) -= tuOtOu But, by hypothesis,

I + -R log " 7

Therefore

o+o+ Ot Ou

and also 02+

OtOu

whether t tends to o or u tends to o.

K(t, u ) = cos +(t, u), 02K _ ~ 0 2 +

tu tsin ,~ ~ -t- o+0+ t cos + 07 gut "

I as t tends to o and ~ log-fir as u tends to o.

I

tu

I tu

whether t tends to o or u tends to o,

whether t tends to o or u tends to o.

Hence f(t, u) is continuous in t, u at (o, o) and, therefore,

02F(o, o) and O=F(o, o) exist and are equal. OxOy OyOx

I ( I -~)1 EXAMVLE. - - For K(t, u) = cos log log-/r- X log , F(x, y) has

differential coefficients at (o, o) which equal o. Proof of ( H ) : Since

KO, . ) = z(t, ~)cos+(t, .), . t ~'K ( [ a 2 Z a+o3+__~ ) ( ~ O+ aZO'$ ~--~+Ou) } f(t, u)-- uo~----tuJ%3~--Z3 ~ cos+-- 0u { 0u 0t' + Z sin+

which, because of the hypothesis, tends to o whether t or u tends to o. Hence f(t, u) is continuous in t, u at (o, o) and, therefore,

02 F(o, o) and c32F(o, o) exist and are equal. OxOy OyOx

EXAMPLE. - - For ! r _!_1 _L

K(t, u)---(log ~-~-)* (log ul--w)' cos I(log tl~-) ' >< (log ul-r)31

F(x, y) has mixed differential coefficients at (o, o) which equal o.

(H)

mixed

Calcut ta , M a r c h io , I9~ 7.

G A N E S H P R A S A D