Q11Mathematics I

Question

If $θ = t^{n} exp (- r^{2} /4 t)$ , find the value of $n$ for which the relation

\frac{1}{r ^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial θ}{\partial r}) = \frac{\partial θ}{\partial t}

is true.

Answer 1

The objective is to rigorously prove the famous identity $B (m, n) = \frac{Γ ( m ) Γ ( n )}{Γ ( m + n )}$ . This proof relies on multiplying two independent Gamma functions, interpreting them geometrically as a double area integral, and changing the coordinate system.

Step 1: The Gamma Function Definitions

We start with the standard Eulerian integral definition of the Gamma function:

Γ (m) = \int_{0}^{\infty} t^{m - 1} e^{- t} d t

To facilitate the transformation to polar coordinates later, we apply a specific algebraic substitution: let $t = x^{2}$ . This means $d t = 2 x d x$ . The limits remain $0$ to $\infty$ . Substituting this yields a new form:

Γ (m) = \int_{0}^{\infty} (x^{2})^{m - 1} e^{- x^{2}} (2 x) d x = 2 \int_{0}^{\infty} x^{2 m - 1} e^{- x^{2}} d x

Using a completely independent dummy variable $y$ , we write the exact same structural formula for $Γ (n)$ :

Γ (n) = 2 \int_{0}^{\infty} y^{2 n - 1} e^{- y^{2}} d y

Step 2: Forming the Double Integral

We now multiply these two independent definite integrals together. Because $x$ and $y$ are completely independent mathematical variables, the product of their integrals equals the double integral over the entire first Cartesian quadrant (since both $x$ and $y$ strictly range from $0$ to $\infty$ ):

Γ (m) Γ (n) = (2 \int_{0}^{\infty} x^{2 m - 1} e^{- x^{2}} d x) (2 \int_{0}^{\infty} y^{2 n - 1} e^{- y^{2}} d y)

Γ (m) Γ (n) = 4 \int_{0}^{\infty} \int_{0}^{\infty} x^{2 m - 1} y^{2 n - 1} e^{- (x^{2} + y^{2})} d x d y

Step 3: Transformation to Polar Coordinates

To evaluate this complex integral, we mathematically transition to polar coordinates. The standard transformation equations are $x = r cos θ$ and $y = r sin θ$ . The differential area element transforms as $d x d y = r d r d θ$ . The first quadrant in Cartesian space corresponds exactly to the radial limits $r \in [0, \infty)$ and the angular limits $θ \in [0, π /2]$ .

We substitute all these components meticulously into the double integral:

Γ (m) Γ (n) = 4 \int_{0}^{π /2} \int_{0}^{\infty} (r cos θ)^{2 m - 1} (r sin θ)^{2 n - 1} e^{- r^{2}} r d r d θ

We group the $r$ terms and $θ$ terms algebraically. The total exponent on $r$ is $(2 m - 1) + (2 n - 1) + 1 = 2 m + 2 n - 1 = 2 (m + n) - 1$ .

Γ (m) Γ (n) = 4 (\int_{0}^{π /2} cos^{2 m - 1} θ sin^{2 n - 1} θ d θ) (\int_{0}^{\infty} r^{2 (m + n) - 1} e^{- r^{2}} d r)

Step 4: Isolating Beta and Gamma Components

We cleverly distribute the constant factor $4$ as $2 \times 2$ to match our earlier derived definitions perfectly:

Γ (m) Γ (n) = (2 \int_{0}^{π /2} sin^{2 n - 1} θ cos^{2 m - 1} θ d θ) \times (2 \int_{0}^{\infty} r^{2 (m + n) - 1} e^{- r^{2}} d r)

We recognize the first bracketed term as the exact, rigorous trigonometric formulation of the Beta function, $B (m, n)$ . We recognize the second bracketed term as the exact squared formulation of the Gamma function we derived in Step 1, but with the parameter $(m + n)$ instead of $m$ .

Γ (m) Γ (n) = B (m, n) \times Γ (m + n)

By simply dividing both sides of the equation by $Γ (m + n)$ , the theorem is completely and flawlessly verified:

B (m, n) = \frac{Γ ( m ) Γ ( n )}{Γ ( m + n )}

Q11Mathematics I

Question

Answer

Step 1: The Gamma Function Definitions

Step 2: Forming the Double Integral

Step 3: Transformation to Polar Coordinates

Step 4: Isolating Beta and Gamma Components