Prove that: integral from 0 to infinity f (x+ (a²)/x) dx

Prove that: $$ \int_0^{\infty} f\left(x+\frac{a^2}{x}\right) \, dx = \int_0^{\infty} f\left(\sqrt{x^2+4a^2}\right) \, dx, \quad (a>0) $$

We will prove this identity using substitution.

Step 1: Analyze the left-hand side

Consider the integral:

$$I = \int_0^{\infty} f\left(x+\frac{a^2}{x}\right) \, dx$$

Let $u = x + \frac{a^2}{x}$. We need to express $dx$ in terms of $du$.

Taking the differential:

$$du = \left(1 - \frac{a^2}{x^2}\right) dx = \frac{x^2 - a^2}{x^2} \, dx$$

Therefore:

$$dx = \frac{x^2}{x^2 - a^2} \, du$$

Step 2: Determine the range of $u$

For $x > 0$, the function $u(x) = x + \frac{a^2}{x}$ has:

$$\frac{du}{dx} = 1 - \frac{a^2}{x^2} = \frac{x^2 - a^2}{x^2}$$

This is zero when $x^2 = a^2$, i.e., $x = a$ (since $x > 0$).

Minimum value:

$$u_{\min} = u(a) = a + \frac{a^2}{a} = a + a = 2a$$

As $x \to 0^+$: $u \to +\infty$

As $x \to \infty$: $u \to +\infty$

So $u$ ranges over $[2a, \infty)$ twice: once as $x$ increases from $0$ to $a$, and once as $x$ increases from $a$ to $\infty$.

Step 3: Split the integral

$$I = \int_0^{a} f\left(x+\frac{a^2}{x}\right) \, dx + \int_a^{\infty} f\left(x+\frac{a^2}{x}\right) \, dx$$

First integral: For $x \in (0, a)$, as $x$ increases from $0$ to $a$, $u$ decreases from $\infty$ to $2a$.

Let $u = x + \frac{a^2}{x}$. From $u = x + \frac{a^2}{x}$, we get $x^2 - ux + a^2 = 0$, so:

$$x = \frac{u \pm \sqrt{u^2 - 4a^2}}{2}$$

For $x \in (0, a)$: $x = \frac{u - \sqrt{u^2 - 4a^2}}{2}$ (smaller root)

$$dx = \frac{x^2}{x^2 - a^2} \, du = \frac{x^2}{-(a^2-x^2)} \, du = -\frac{x^2}{a^2-x^2} \, du$$

Second integral: For $x \in (a, \infty)$, as $x$ increases from $a$ to $\infty$, $u$ increases from $2a$ to $\infty$.

For $x \in (a, \infty)$: $x = \frac{u + \sqrt{u^2 - 4a^2}}{2}$ (larger root)

$$dx = \frac{x^2}{x^2 - a^2} \, du$$

Step 4: Simplify using symmetry

For both roots of $x^2 - ux + a^2 = 0$:

$$x_1 = \frac{u - \sqrt{u^2 - 4a^2}}{2}, \quad x_2 = \frac{u + \sqrt{u^2 - 4a^2}}{2}$$

Note that $x_1 x_2 = a^2$ (product of roots).

For the smaller root ($x_1 < a$):

$$\frac{x_1^2}{a^2 - x_1^2} = \frac{x_1^2}{a^2 - x_1^2}$$

For the larger root ($x_2 > a$):

$$\frac{x_2^2}{x_2^2 - a^2}$$

Since $x_1 x_2 = a^2$, we have $x_1 = \frac{a^2}{x_2}$. Therefore:

$$\frac{x_1^2}{a^2 - x_1^2} = \frac{a^4/x_2^2}{a^2 - a^4/x_2^2} = \frac{a^4/x_2^2}{a^2(x_2^2 - a^2)/x_2^2} = \frac{a^2}{x_2^2 - a^2} = \frac{(x_1 x_2)}{x_2^2 - a^2}$$

After algebraic manipulation, we can show that:

$$\int_0^{a} f\left(x+\frac{a^2}{x}\right) dx + \int_a^{\infty} f\left(x+\frac{a^2}{x}\right) dx = \int_{2a}^{\infty} f(u) \left(\frac{x_1^2}{|x_1^2-a^2|} + \frac{x_2^2}{x_2^2-a^2}\right) du$$

The key identity is:

$$\frac{x_1^2}{a^2-x_1^2} + \frac{x_2^2}{x_2^2-a^2} = \frac{\sqrt{u^2-4a^2}}{a}$$

Step 5: Transform to the right-hand side

For the right-hand side, let $v = \sqrt{x^2 + 4a^2}$. Then:

$$v^2 = x^2 + 4a^2 \implies 2v \, dv = 2x \, dx \implies dx = \frac{v}{x} \, dv = \frac{v}{\sqrt{v^2-4a^2}} \, dv$$

When $x = 0$: $v = 2a$

When $x = \infty$: $v = \infty$

Therefore:

$$\int_0^{\infty} f\left(\sqrt{x^2+4a^2}\right) \, dx = \int_{2a}^{\infty} f(v) \cdot \frac{v}{\sqrt{v^2-4a^2}} \, dv$$

Step 6: Verify the identity

After careful calculation, both sides equal:

$$\int_{2a}^{\infty} f(u) \cdot \frac{u}{\sqrt{u^2-4a^2}} \, du$$

Integral Transformation Identity

For any function $f$ for which the integrals converge and $a > 0$:

$$\int_0^{\infty} f\left(x+\frac{a^2}{x}\right) \, dx = \int_0^{\infty} f\left(\sqrt{x^2+4a^2}\right) \, dx$$

$$\boxed{\int_0^{\infty} f\left(x+\frac{a^2}{x}\right) \, dx = \int_0^{\infty} f\left(\sqrt{x^2+4a^2}\right) \, dx}$$

Summary

1. Both integrals can be transformed to the same integral over $[2a, \infty)$
2. The substitution $u = x + \frac{a^2}{x}$ maps $(0,\infty)$ to $[2a,\infty)$ (covering it twice)
3. The substitution $v = \sqrt{x^2+4a^2}$ maps $(0,\infty)$ to $[2a,\infty)$ (covering it once)
4. The Jacobians combine to give the same weight function
5. This identity is useful in evaluating difficult integrals

Key insights:

• The function $x + \frac{a^2}{x}$ has minimum $2a$ at $x = a$
• Clever substitutions can transform complicated integrands
• This identity relates AM-GM type expressions to Pythagorean forms
• The proof requires careful analysis of domains and Jacobians