Di erentiation of integrals and Leibniz rule - University of Central ...

Differentiation of integrals and Leibniz rule

? We are interested in solving problems of the type

d v (x) f (x, t)dt

dx u(x)

? Notice in addition to the limits depending on x, the function f (x, t) also depends on x ? First lets take the case where the limits are constants (u(x) = b and v (x) = a) ? In most instances we can take the derivative inside the integral (as a partial derivative)

db

b f (x, t)

f (x, t)dt =

dt

dx a

a x

Patrick K. Schelling

Introduction to Theoretical Methods

Example

? Find

0

tne

-kt 2

dt

for

n

odd

? First solve for n = 1,

I=

te-kt2 dt

0

? Use u = kt2 and du = 2ktdt, so

I=

te-kt2 dt =

1

e-udu =

1

0

2k 0

2k

?

Next

notice

dI dk

=

0

-t

3e

-kt

2

dt

=

-

1 2k

2

? We can take successive derivatives with respect to k, so we find

t2n+1e-kt2 dt

0

=

n! 2k n+1

Patrick K. Schelling

Introduction to Theoretical Methods

Another example... the even n case

? The even n case can be done of

0

t

n

e

-kt

2

dt

? Take for I in this case (writing in terms of x instead of t)

? We can write for I 2,

I=

e-kx2 dx

-

I2 =

e -k (x 2 +y 2 ) dxdy

- -

? Make a change of variables to polar coordinates, x = r cos , y = r sin , and x2 + y 2 = r 2

? The dxdy in the integrand becomes rdrd (we will explore this

more carefully in the next chapter)

Patrick K. Schelling

Introduction to Theoretical Methods

Example: continued

? Then in polar coordinates we solve the I 2 integral,

I2 =

2

e-kr2 rdrd

=

00

k

? So we find I =

k

? To find

0

x

2

e

-kx

2

dx

,

we

take

-

1 2

dI dk

=

1 2

1 1/2 2 k3/2

? For

0

x

4e

-kx

2

dx

=

1 d2I 2 dk2

=

1 2

1 2

3 1/2 2 k5/2

0

x 2ne-kx2 dx

=

1/2 2n+1

(2n - 1)!! k (2n+1)/2

Patrick K. Schelling

Introduction to Theoretical Methods

Continuing Leibniz rule...

? Remember we wanted to solve problems like

d v (x) f (x, t)dt

dx u(x)

? So far have only considered the case u(x) = a and v (x) = b were just constants ? Let us now look at the problem when f (x, t) is only a function of t, f (t)

dI d v(x)

=

f (t)dt

dx dx u(x)

? We see I (u, v ) and also u(x), v (x) are functions of x, so

dI I du I dv

=

+

dx u dx v dx

?

Let's

see

how

we

find

I u

and

I v

Patrick K. Schelling

Introduction to Theoretical Methods

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