n! for any n ∈ N

(1)

For x > 0, we set

Γ(x) = Z ∞

0

t^x−1e^−tdt.

We can show that Γ(x) is convergent. Hence Γ defines a function from (0, ∞) to R. It is not difficult for us to show that

(1) Γ(x + 1) = xΓ(x) for any x > 0;

(2) Γ(1) = 1.

By induction, Γ(n + 1) = n! for any n ∈ N. In other words, (1.1)

Z ∞ 0

tⁿe^−tdt = n!, n ∈ N.

We can compute the integral (1.1) via other method. For each s > 0, we define F (s) =

Z ∞ 0

e^−stdt = 1 s. Let us assuming that the following formula holds

F⁰(s) = Z ∞

0

∂

∂s e^−st dt.

By F⁰(s) = −1/s², we find

Z ∞ 0

te^−stdt = 1 s². Inductively, we obtain that

Z ∞ 0

tⁿe^−stdt = n!

sⁿ⁺¹

for any natural number n ∈ N. Plugging s = 1, we obtain (1.1). In this note, we are going to learn techniques of evaluating definite integrals or improper integrals via the method of differentiation. Let us take a look at other examples.

Example 1.1. Evaluate the integral Z ∞

0

e^−txsin x

x dx for t > 0.

We know that | sin x/x| ≤ 1 for any x > 0. Then 0 ≤ e^−tx

sin x x

≤ e^−tx

for any x > 0 if t > 0. Since Z ∞

0

e^−txdx is convergent, by comparison test, the above integral is absolutely convergent if t > 0. Therefore the integral defines a function from (0, ∞) to R.

We define this function by F (t). Let us assume that F is differentiable. This fact can be verified and is left to the reader. Let us assume further that the following formula holds:

F⁰(t) = Z ∞

0

∂

∂t

e^−txsin x x

dx = − Z ∞

0

e^−txsin xdx.

By elementary calculus, Z

e^axsin bxdx = a sin bx − b cos bx a²+ b² + C.

1

(2)

Hence we obtain that

Z ∞ 0

e^−txsin xdx = 1 1 + t².

This implies that F⁰(t) = −1/(1 + t²) and therefore F (t) = C − tan⁻¹t for t > 0. On the other hand,

|F (t)| ≤ Z ∞

0

e^−tx

sin x x

dx ≤ Z ∞

0

e^−txdx = 1 t. We find that lim

t→∞F (t) = 0. Therefore

t→∞lim F (t) = lim

t→∞ C − tan⁻¹t = C −π 2. Thus C = π/2. Furthermore, F (0) = π/2 which gives

Z ∞ 0

sin x

x dx = π 2.

Example 1.2. (The Gaussian integral and the its moments) For any nonnegative integer n, evaluate the integral cn=

Z ∞

−∞

xⁿe⁻¹²^x²dx.

We can prove that the integrals converges absolutely via the limit comparison test. The convergence of the integrals are left to the readers. Let us define

F (t) = Z ∞

−∞

e^txe⁻¹²^x²dx for any t. By taking the derivatives of F, we find

F^(k)(t) = Z ∞

−∞

x^ke^txe⁻¹²^x²dx.

Then cn= F⁽ⁿ⁾(0) for any n ≥ 0. When n = 2k + 1 is odd, the function x 7→ x^2k+1e⁻¹²^x² is an odd function on R. The integral

Z ∞

−∞

x^2k+1e⁻¹²^x²dx is zero, i.e. c_2k+1= 0. Now we only need to compute c_2k for any k ≥ 0. By completing the square we know

tx −x² 2 = −1

2(x + t) + t² 2. Therefore

F (t) = Z ∞

−∞

e⁻¹²^(x+t)+^t2² dx = e^t2² Z ∞

−∞

e⁻¹²^(x+t)²dx.

Since the integral is invariant under the change of variable change of variable u = x + t, F (t) = e¹²^t²c₀.

Using the Taylor expansion of F (t), we know that F (t) =

∞

X

n=0

c0

2ⁿn!t²ⁿ, t ∈ R.

By uniqueness of the Taylor expansion, F²ⁿ(0)

(2n)! = c₀

2ⁿn!, n ≥ 0.

(3)

We find that

c2n= Z ∞

−∞

x²ⁿe⁻¹²^x²dx = (2n)!

2ⁿn!c0. Now we only need to compute c0. To do this, we set

G(t) = Z ∞

0

e⁻¹²^t²^(1+x²⁾ 1 + x² dx.

Differentiating G with respect to t, one has G⁰(t) =

Z ∞

−∞

e⁻¹²^t²^(1+x²⁾· (−t)dx.

Making a change of variable u = xt, we find Z ∞

0

e⁻¹²^t²^(1+x²⁾· (−t)dx = − Z ∞

0

e⁻¹²^t²⁻¹²^u²du = −e⁻¹²^t² Z ∞

−∞

e⁻¹²^u²du.

Hence G⁰(t) = −e⁻¹²^t²c₀/2. By fundamental Theorem of calculus, G(b) − G(0) = −c₀

2 Z b

0

e⁻¹²^t²dt.

On the other hand, lim

b→∞G(b) = 0 and G(0) = π/2. We find that

−π

2 = −c0

2 Z ∞

0

e⁻¹²^t²dt = −c²₀ 4. This implies that c0 =√

2π.

Let us study when these calculation make senses.

Lemma 1.1. Let R = [a, b] × [c, d] be a compact interval in R² and h : R → R be a continuous function. Define H : [a, b] → R by

H(x) = Z d

c

h(x, y)dy.

Then H : [a, b] → R is uniformly continuous; hence

x→tlimH(x) = H(t) ⇐⇒ lim

x→t

Z d c

h(x, y)dy = Z d

c

x→tlimh(x, y) dy.

Proof. Since R is compact and h : R → R is continuous, h is uniformly continuous. For any

> 0, we can find δ > 0 so that

|h(x, y) − h(x⁰, y⁰)| < d − c

whenever k(x, y) − (x⁰, y⁰)k < δ. If |x − x⁰| < δ, then for any y ∈ [c, d], k(x, y) − (x⁰, y)k =

|x − x⁰| < δ. In this case, |h(x, y) − h(x⁰, y)| < /(d − c). If |x − x⁰| < δ,

|H(x) − H(x⁰)| ≤ Z _d

c

|h(x, y) − h(x⁰, y)|dy <

Z _d

c

d − cdy = .

We complete the proof of our assertion.

(4)

Theorem 1.1. Let U be an open subset of R² and f : U → R be a function such that f_x exist and continuous on U. For any compact subinterval [a, b] × [c, d] of U, we define F : [a, b] → R by

F (x) = Z d

c

f (x, y)dy.

Then F ∈ C¹[a, b] such that F⁰(x) = Z d

c

fx(x, y)dy.

Proof. Let x be an interior point of [a, b]. Choose δ > 0 so that (x − δ, x + δ) is contained in [a, b]. For 0 < |h| < δ, we compute the difference quotient

F (x + h) − F (x)

h =

Z d c

f (x + h, y) − f (x, y)

h dy.

By the mean value theorem, there exists θ ∈ [0, 1] so that f (x + h, y) − f (x, y) = f_x(x + θh, y)h.

When 0 < |h| < δ,

F (x + h) − F (x)

h =

Z d c

fx(x + θh, y)dy.

Since f_x is continuous on U, by Lemma 1.1, we know

h→0lim

F (x + h) − F (x)

h = lim

h→0

Z d c

fx(x + θh, y)dy = Z d

c

h→0limfx(x + θh, y)dy = Z d

c

fx(x, y)dy.

Therefore F⁰(x) exists and

F⁰(x) = Z d

c

fx(x, y)dy.

We leave to the reader to verify that the left derivative of F at b and right derivative of F at a both exist. By Lemma 1.1 again, F⁰(x) is continuous on [a, b]. We complete the proof

of our result.

Corollary 1.1. Let g ∈ C²(U ) where U is an open subset of R². Then g_xy = g_yx on U.

Proof. We verify that gxy(x0, y0) = gyx(x0, y0) for any (x0, y0) ∈ U. We choose a compact interval R = [a, b] × [c, d] containing (x₀, y₀). Let f : U → R be the function f (x, y) = gy(x, y). By the fundamentla Theorem of calculus,

g(x, y) = g(x, d) + Z y

d

f (x, t)dt for any (x, y) ∈ R. By Theorem 1.1, we find that

g_x(x, y) = g_x(x, d) + Z y

d

f_x(x, t)dt, (x, y) ∈ R.

It follows from the fundamental Theorem of calculus again, (differentiating the above equa- tion with respect to y)

g_xy(x, y) = f_x(x, y) = g_yx(x, y)

for any (x, y) ∈ R. We complete the proof of our result.

(5)

In calculus, we have learned the following formula d

dx

Z ψ2(x) ψ1(x)

f (t)dt = f (ψ2(x))ψ⁰₂(x) − f (ψ1(x))ψ⁰₁(x), where f, ψ₁, ψ₂ are some C¹-functions. Now we would like to investigate

d dx

Z ψ2(x) ψ1(x)

f (x, y)dy.

More precisely, let f ∈ C²(U ) and ψ₁, ψ₂ : [a, b] → R be two C¹ functions on [a, b] such that a < ψ₁(x) < ψ₂(x) < b

for any x ∈ [a, b]. Define F : [a, b] → R by F (x) =

Z ψ2(x) ψ1(x)

f (x, y)dy.

Then we would like to compute F⁰(x). To do this, we set G(u, v, x) =

Z v u

f (x, y)dy,

where (u, v) ∈ [c, d]² and x ∈ [a, b]. The function F (x) can be rewritten as F (x) = G(ψ1(x), ψ2(x), x), x ∈ [a, b].

By chain rule,

F⁰(x) = Gu(ψ1(x), ψ2(x), x)ψ⁰₁(x) + Gv(ψ1(x), ψ2(x), x)ψ₂⁰(x) + Gx(ψ1(x), ψ2(x), x).

It is not difficult to verify that

Gu(u, v, x) = −f (x, u), Gv(u, v, x) = f (x, v), Gx(u, v, x) = Z v

u

fx(x, y)dy.

We conclude that

F⁰(x) = f (x, ψ₂(x))ψ⁰₂(x) − f (x, ψ₁(x))ψ⁰₁(x) +

Z ψ2(x) ψ1(x)

f_x(x, y)dy, i.e.

d dx

Z ψ2(x) ψ1(x)

f (x, y)dy = f (x, ψ2(x))ψ₂⁰(x) − f (x, ψ1(x))ψ₁⁰(x) +

Z ψ2(x) ψ1(x)

fx(x, y)dy, for any x ∈ [a, b].

Example 1.3. Evaluate Z 1

0

(x − 1)x^k

log x dx for k > −1.

We leave it to the reader to verify that (1) lim

k→∞kF (k) = 1 (2) F⁰(k) = 1

k + 2 − 1 k + 1.

From here, you will be able to determine F (k).

(6)

Corollary 1.2. Let f ∈ C([a, b] × [c, d]) and I =

Z b a

Z d c

f (x, y)dy

dx and J = Z d

c

Z b a

f (x, y)dx

dy.

Then I = J.

Proof. Let us define two functions u, v : [a, b] × [c, d] → R by v(x, y) =

Z y c

f (x, s)ds u(x, y) =

Z x a

v(t, y)dt.

Then u(b, d) = I and vy = f (x, y). Differentiating u with respect to y, we find uy(x, y) =

Z x a

vy(t, y)dt = Z x

a

f (t, y)dt.

On the other hand, by the fundamental theorem of calculus, u(x, y) = u(x, c) +

Z y c

u_y(x, s)ds = Z y

c

Z x a

f (t, s)dt

ds

This implies that u(b, d) = J. We conclude that I = J.

Now let us assume that I × J is an unbounded closed interval in R² where I and J are intervals of R such that I or J could be unbounded; in other words, I and J could be one of the following integrals (−∞, ∞) or [a, ∞) or (−∞, b] for some a, b ∈ R. Assume that f is a function defined on an open subset U of R² containing I × J so that f and f_x are continuous on U. We define

F : I → R, F (x) = Z

J

f (x, y)dy.

We would like to investigate the differentiability of F.

Theorem 1.2. Suppose that there exist nonnegative A, B : J → R such that (1) |f (x, y)| ≤ A(y) and |f_x(x, y)| ≤ B(y) for any (x, y) ∈ I × J and (2)

Z

J

A(y)dy and Z

J

B(y)dy are both convergent.

Then F is differentiable and

F⁰(x) = Z

J

fx(x, y)dy.