Differentiation Under the Integral Sign brilliant PDF

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3/14/2021

Differentiation Under the Integral Sign | Brilliant Math & Science Wiki

Dierentiation Under the Integral Sign Dierentiation under the integral sign is an operation in calculus used to evaluate certain integrals. Under fairly loose conditions on the function being integrated, dierentiation under the integral sign allows one to interchange the order o integration and dierentiation. In its simplest form, called the Leibniz integral rule, dierentiation under the integral sig makes the following equation valid under light assumptions on f: b b d ∂ ∫ f (x, t) dt = ∫ f (x, t) dt. dx a a ∂x

Many integrals that would otherwise be impossible or require significantly more complex methods can be solved by this approach.

Contents General Form Examples

General Form

f (x, t) is a continuous and continuously dierentiable (i.e., partial derivatives exist and are themselves continuous) function and the limits of integration a(x)an are continuous and continuously dierentiable functions of x, then

The most general form of dierentiation under the integral sign states that: if

b(x) b(x) ∂ d ′ ′ ∫ f (x, t) dt = f (x, b(x)) ⋅ b (x) − f (x, a(x)) ⋅ a (x) + ∫ f (x, t) dt. dx a(x) a(x) ∂x

In the case where

a(x) and b(x) are constant functions, this formula reduces to the simpler form b b ∂ d f (x, t) dt. ∫ f (x, t) dt = ∫ dx a a ∂x

This simpler statement is known as Leibniz integral rule.

Examples Generally, one uses dierentiation under the integral sign to evaluate integrals that can be thought of as belonging to so family of integrals parameterized by a real variable. To beer understand this statement, consider the following example EXAMPLE

Compute the definite integral

∫

1 0

t3 − 1 dt. ln t

This integral appears resistant to standard integration techniques such as integration by parts, u-substitution, etc. We w like to use dierentiation under the integral sign to compute it. https://brilliant.org/wiki/differentiate-through-the-integral/

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Differentiation Under the Integral Sign | Brilliant Math & Science Wiki

How can we choose a function to dierentiate under the integral sign? The appearance of

ln t in the denominator of th

integrand is quite unwelcome, and we would like to get rid of it. Thankfully, we know

d x t = t x ln t, dx so dierentiating the numerator with respect to the exponent seems to be what we'd like to do. Accordingly, we define a function

g(x) = ∫

1

tx − 1 dt. ln t

0

g(3). Observe that the given integral has been recast as member of family of definite integrals g(x) indexed by the variable x.

In this notation, the integral we wish to evaluate is

By Leibniz integral rule, we compute

g′ (x) = ∫ It follows that g(x)

1 0

∣1 1 x t ln t 1 ∂ tx − 1 tx+1 ∣ dt = dt = ∫ . ∣ = ln t x+1 ∂x ln t x + 1∣ 0 ∣0

= ln ∣x + 1∣ + C for some constant C .

= 0, so 0 = g(0) = ln 1 + C = C . Hence, g(x) = ln ∣x + 1∣ for all x such that the integral exists. In particular, g(3) = ln 4 = 2 ln 2 . □ To determine C , note that g(0)

In the example, part of the integrand was replaced with a variable and the resultant function was studied using dierent under the integral sign. This is a good illustration of the problem-solving principle: if stuck on a specific problem, try solv more general problem. TRY IT YOURSELF

Compute the definite integral

49! 5051

1

∫ (x ln x)50 dx.

51! 5050

0

50! 5151 50! 4951

Another example illustrates the power of this technique in its general form, as one may use it to compute the Gaussian in EXAMPLE

Compute the definite integral

∫

https://brilliant.org/wiki/differentiate-through-the-integral/

∞

2

e −x

/2

dx.

0

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Differentiation Under the Integral Sign | Brilliant Math & Science Wiki

Define a function t

2

g(t) = (∫ e −x

dx) .

2 /2

0

Our goal is to compute g(∞) and then take its square root. Dierentiating with respect to t gives t

g′ (t) = 2 ⋅ (∫ e−x 0

Make the change of variables u

2

/2

dx ) ⋅ (

t t t 2 2 d 2 2 2 ∫ e−x /2 dx) = 2e −t /2 ∫ e −x /2 dx = 2 ∫ e−(t +x )/2 dx. dt 0 0 0

= x/t, so that the integral transforms to g′ (t) = 2 ∫

1

2

te −(1+u

)t2 /2

du.

0

Now, the integrand has a closed-form antiderivative with respect to t:

g (t) = −2 ∫

1

′

0

2

2

2

2

∂ e−(1+u )t /2 d 1 e−(1+u )t /2 −2 du. du = ∫ dt 0 ∂t 1 + u2 1 + u2

Set

h(t) = ∫ Then by the above calculation, g′ (t)

1 0

2

2

e−(1+x )t /2 dx. 1 + x2

= −2h ′(t) , so g(t) = −2h(t) + C . To determine C , take t → 0 in the equation;

g(0) = 0 and h(0) = ∫

1 0

∣1 π 1 −1 ∣ dx = tan x∣ = , 2 ∣ 4 1+x ∣0

it follows that 0

= −π/2 + C ⟹ C = π/2 .

Finally, taking t

→ ∞ , we conclude g(∞) = −2h(∞) + π/2 = π/2 . Thus, ∫

∞

e−x

2

/2

dx =

0

π .□ 2

EXAMPLE

Let a, b, c be real with c

> 0 . Show that ∫

Let a, b, c be real with c

∞

2

e−cx erf(ax + b) dx =

−∞

b c π erf ( ) c a2 + c

> 0 . Define I(a, b, c) = ∫

https://brilliant.org/wiki/differentiate-through-the-integral/

∞

2

e −cx erf(ax + b)dx −∞ 3/5

3/14/2021

Differentiation Under the Integral Sign | Brilliant Math & Science Wiki

If we set u

=

c x then we find I(a, b, c) = ∫

So we will focus on determining

∞ −∞

a 1 a du u + b) = I ( , b, 1 ) c c c c

e−u erf ( 2

J(a, b) = I(a, b, 1). Recall that erf x =

2 π

∫

x

2

e−t dt

⟹

erf ′(x) =

0

and so, dierentiating under the integral sign with respect to

2 −x2 e π

a we find

∞ ∞ 2 ∂J 2 2 2 ∂ ∫ xe−x e−(ax+b) dx =∫ e −x erf(ax + b)dx = π −∞ ∂a ∂a −∞

Completing the square in the exponent we have

x 2 + (ax + b)2 = ( a2 + 1) x2 + 2abx + b 2 = ( a2 + 1) ( x +

2

b2 ab ) + a2 + 1 a2 + 1

Hence ∞ ∂J −b2 ab 2 = exp ( 2 ) ∫ x exp [− (a2 + 1) (x + 2 ) ] dx π ∂a a +1 a +1 −∞ 2

∞ ab 2 −b2 ) exp [− (a 2 + 1 )v 2 ]dv [v = x + ab/ (a 2+ 1 )] ) ∫ (v − 2 exp ( 2 a +1 a + 1 π −∞ ∞ 2 2 −b2 ab 2 )∫ = exp [− (a 2 + 1 )v 2 ]dv [usingoddnessofve −(a +1)v ] ) exp ( 2 (− 2 π a +1 a +1 −∞ π ab 2 −b2 ) exp ( 2 (− 2 = [(i)] ) 2 a +1 a +1 π a +1 −2ab −b2 ( ) = exp a2 + 1 ( a2 + 1)3/2

=

for (i)

∫ Thus, noting J(a, b)

∞

2

e−x dx =

−∞

1 and Γ ( ) = 2

π

→ 0 as a → −∞ , we finally have J(a, b) = −2b ∫

We will set u2

π

a −∞

x ( x2 + 1)3/2

exp (

−b2 ) dx x2 + 1

= b2 /(1 + x 2) so that 2udu =

−2b 2x dx = ( 1 + x2 )2

u −2b 2x −2b 2x 1 ( dx) d x = b ( 1 + x2 )3/2 1 + x2 ( 1 + x2 )3/2

and

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Differentiation Under the Integral Sign | Brilliant Math & Science Wiki

du =

−bxdx ( 1 + x2 )3/2

Hence

J(a, b) = 2 ∫

b/ 1+a2

exp (−u 2 ) du =

0

π erf (

b ) 1 + a2

and

I(a, b, c) =

1 c

J(

a , b) = c

π erf ( c

b 1 + a2/c

)=

π b c ) erf ( c c + a2

as required One should also note counterexamples, for which this technique does not work. For instance, suppose one aempts to e

∫ by making the variable change

∞ 0

sin x dx x

u = x/t for some t  = 0. Then, ∫

∞ 0

∞ sin tu sin x du = g(t). dx = ∫ u x 0

Dierentiating under the integral sign yields

0 = g ′(t) = ∫ which is absurd. The problem is that the function

∞

cos tu du, 0

f (x, t) = sin tx/x is not continuously dierentiable (consider ∂f /∂ t

x = 0), which was required in the assumptions set forth above. TRY IT YOURSELF

Compute the definite integral:

4π

∫

2π

ecos θ cos(sin θ) dθ.

2π

0

π

Hint: Consider the function

f (t) = ∫

2π

et cos θ cos(t sin θ) dθ

0

0

and use dierentiation under the integral sign.

Cite as: Dierentiation Under the Integral Sign. Brilliant.org. Retrieved 20:06, March 14, 2021, from hps://brilliant.org/wiki/dierentiate-through-the-integral/

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