Calculate the integrals: J_1=integral from 0 to infinity

Calculate the integrals: $$ J_1=\int_0^{\infty} \frac{dx}{(x^2+a^2)(x^2+b^2)} \quad \text{and} \quad J_2=\int_0^{\infty} \frac{x^2 \, dx}{(x^2+a^2)(x^2+b^2)} $$ where $a \neq 0$ and $a + b \neq 0$.

Part 1: Calculate $J_1$

Step 1: Partial fraction decomposition

We decompose:

$$\frac{1}{(x^2+a^2)(x^2+b^2)} = \frac{A}{x^2+a^2} + \frac{B}{x^2+b^2}$$

Multiplying both sides by $(x^2+a^2)(x^2+b^2)$:

$$1 = A(x^2+b^2) + B(x^2+a^2)$$

$$1 = (A+B)x^2 + (Ab^2 + Ba^2)$$

Comparing coefficients:

• Coefficient of $x^2$: $A + B = 0 \implies B = -A$
• Constant term: $Ab^2 + Ba^2 = 1 \implies Ab^2 - Aa^2 = 1$

From the second equation:

$$A(b^2 - a^2) = 1 \implies A = \frac{1}{b^2-a^2}$$

Therefore:

$$A = \frac{1}{b^2-a^2}, \quad B = -\frac{1}{b^2-a^2} = \frac{1}{a^2-b^2}$$

Step 2: Integrate

$$J_1 = \int_0^{\infty} \left(\frac{1}{b^2-a^2} \cdot \frac{1}{x^2+a^2} + \frac{1}{a^2-b^2} \cdot \frac{1}{x^2+b^2}\right) dx$$

$$= \frac{1}{b^2-a^2} \int_0^{\infty} \frac{dx}{x^2+a^2} - \frac{1}{b^2-a^2} \int_0^{\infty} \frac{dx}{x^2+b^2}$$

Using $\int \frac{dx}{x^2+c^2} = \frac{1}{c}\arctan\frac{x}{c} + C$:

$$\int_0^{\infty} \frac{dx}{x^2+a^2} = \frac{1}{|a|}\left[\arctan\frac{x}{|a|}\right]_0^{\infty} = \frac{1}{|a|} \cdot \frac{\pi}{2}$$

Similarly:

$$\int_0^{\infty} \frac{dx}{x^2+b^2} = \frac{\pi}{2|b|}$$

Therefore:

$$J_1 = \frac{1}{b^2-a^2}\left(\frac{\pi}{2|a|} - \frac{\pi}{2|b|}\right) = \frac{\pi}{2(b^2-a^2)}\left(\frac{1}{|a|} - \frac{1}{|b|}\right)$$

$$= \frac{\pi}{2(b^2-a^2)} \cdot \frac{|b|-|a|}{|a||b|}$$

For $a, b > 0$:

$$\boxed{J_1 = \frac{\pi}{2(b^2-a^2)} \cdot \frac{b-a}{ab} = \frac{\pi(b-a)}{2ab(b^2-a^2)} = \frac{\pi}{2ab(b+a)}}$$

Part 2: Calculate $J_2$

Step 1: Partial fraction decomposition for numerator with $x^2$

We decompose:

$$\frac{x^2}{(x^2+a^2)(x^2+b^2)} = \frac{Cx+D}{x^2+a^2} + \frac{Ex+F}{x^2+b^2}$$

By symmetry (integrand is even), $C = E = 0$. So:

$$\frac{x^2}{(x^2+a^2)(x^2+b^2)} = \frac{D}{x^2+a^2} + \frac{F}{x^2+b^2}$$

Multiplying by $(x^2+a^2)(x^2+b^2)$:

$$x^2 = D(x^2+b^2) + F(x^2+a^2)$$

$$x^2 = (D+F)x^2 + (Db^2 + Fa^2)$$

Comparing coefficients:

• Coefficient of $x^2$: $D + F = 1$
• Constant term: $Db^2 + Fa^2 = 0$

From the first equation: $F = 1 - D$

Substituting into the second:

$$Db^2 + (1-D)a^2 = 0 \implies Db^2 + a^2 - Da^2 = 0 \implies D(b^2-a^2) = -a^2$$

$$D = \frac{-a^2}{b^2-a^2} = \frac{a^2}{a^2-b^2}$$

$$F = 1 - D = 1 - \frac{a^2}{a^2-b^2} = \frac{a^2-b^2-a^2}{a^2-b^2} = \frac{-b^2}{a^2-b^2} = \frac{b^2}{b^2-a^2}$$

Step 2: Integrate

$$J_2 = \int_0^{\infty} \left(\frac{a^2}{a^2-b^2} \cdot \frac{1}{x^2+a^2} + \frac{b^2}{b^2-a^2} \cdot \frac{1}{x^2+b^2}\right) dx$$

$$= \frac{a^2}{a^2-b^2} \cdot \frac{\pi}{2|a|} + \frac{b^2}{b^2-a^2} \cdot \frac{\pi}{2|b|}$$

For $a, b > 0$:

$$= \frac{\pi}{2(a^2-b^2)} \cdot a + \frac{\pi}{2(b^2-a^2)} \cdot b$$

$$= \frac{\pi a}{2(a^2-b^2)} - \frac{\pi b}{2(a^2-b^2)} = \frac{\pi(a-b)}{2(a^2-b^2)} = \frac{\pi(a-b)}{2(a-b)(a+b)}$$

$$\boxed{J_2 = \frac{\pi}{2(a+b)}}$$

Verification: $J_1 + J_2$

Notice that:

$$\frac{1}{(x^2+a^2)(x^2+b^2)} + \frac{x^2}{(x^2+a^2)(x^2+b^2)} = \frac{1+x^2}{(x^2+a^2)(x^2+b^2)}$$

Therefore: $J_1 + J_2$ should simplify nicely.

$$J_1 + J_2 = \frac{\pi}{2ab(a+b)} + \frac{\pi}{2(a+b)} = \frac{\pi}{2(a+b)}\left(\frac{1}{ab} + 1\right) = \frac{\pi}{2(a+b)} \cdot \frac{1+ab}{ab}$$

Summary of Results

For $a, b > 0$ with $a \neq b$:

$$\boxed{J_1 = \int_0^{\infty} \frac{dx}{(x^2+a^2)(x^2+b^2)} = \frac{\pi}{2ab(a+b)}}$$

$$\boxed{J_2 = \int_0^{\infty} \frac{x^2 \, dx}{(x^2+a^2)(x^2+b^2)} = \frac{\pi}{2(a+b)}}$$

Special relationship:

$$\frac{J_2}{J_1} = ab$$

Key observations:

• $J_1$ depends on both $a$ and $b$ in the denominator
• $J_2$ is independent of $ab$, depending only on $a+b$
• Both integrals converge due to $x^{-4}$ behavior at infinity
• Partial fractions is the essential technique
• The arctangent integral $\int_0^{\infty} \frac{dx}{x^2+c^2} = \frac{\pi}{2c}$ is fundamental