Mathematical Methods for Physics and Engineering : A Comprehensive Guide

INTEGRAL TRANSFORMS

equals zero. This leads immediately to two further useful results:
∫b

−a

δ(t)dt= 1 for alla, b > 0 (13.13)

and
∫
δ(t−a)dt=1, (13.14)

provided the range of integration includest=a.

Equation (13.12) can be used to derive further useful properties of the Dirac

δ-function:

δ(t)=δ(−t), (13.15)

δ(at)=

1

|a|

δ(t), (13.16)

tδ(t)=0. (13.17)

Prove thatδ(bt)=δ(t)/|b|.

Let us first consider the case whereb>0. It follows that
∫∞

−∞

f(t)δ(bt)dt=

∫∞

−∞

f

(

t′

b

)

δ(t′)

dt′

b

=

1

b

f(0) =

1

b

∫∞

−∞

f(t)δ(t)dt,

where we have made the substitutiont′=bt.Butf(t) is arbitrary and so we immediately
see thatδ(bt)=δ(t)/b=δ(t)/|b|forb>0.
Now consider the case whereb=−c<0. It follows that
∫∞

−∞

f(t)δ(bt)dt=

∫−∞

∞

f

(

t′

−c

)

δ(t′)

(

dt′

−c

)

=

∫∞

−∞

1

c

f

(

t′

−c

)

δ(t′)dt′

=

1

c

f(0) =

1

|b|

f(0) =

1

|b|

∫∞

−∞

f(t)δ(t)dt,

where we have made the substitutiont′=bt=−ct.Butf(t) is arbitrary and so

δ(bt)=

1

|b|

δ(t),

for allb, which establishes the result.

Furthermore, by considering an integral of the form

∫

f(t)δ(h(t))dt,

and making a change of variables toz=h(t), we may show that

δ(h(t)) =

∑

i

δ(t−ti)

|h′(ti)|

, (13.18)

where thetiare those values oftfor whichh(t)=0andh′(t) stands fordh/dt.