Find a closed form for the following integral

To evaluate the integral:

[
I = \int_{0}^{\infty} \frac{(\ln x)^2}{(1 + x)^2} , dx
]

we can use the method of differentiation under the integral sign. Here’s a step-by-step approach:

Step 1: Define a Parameterized Integral

Consider the more general integral:

[
I(k) = \int_{0}^{\infty} \frac{x^k}{(1 + x)^2} , dx
]

Our goal is to find ( I”(0) ), the second derivative of ( I(k) ) with respect to ( k ) evaluated at ( k = 0 ), since:

[
I = \left. \frac{d^2}{dk^2} I(k) \right|_{k=0}
]

Step 2: Evaluate ( I(k) )

We can evaluate ( I(k) ) using the Beta function, which is related to the Gamma function:

[
I(k) = \int_{0}^{\infty} \frac{x^k}{(1 + x)^2} , dx = \frac{\Gamma(k + 1) \Gamma(1 – k)}{\Gamma(2)}
]

Using the reflection formula for the Gamma function:

[
\Gamma(k + 1) \Gamma(1 – k) = \frac{\pi k}{\sin(\pi k)}
]

Thus,

[
I(k) = \frac{\pi k}{\sin(\pi k)}
]

Step 3: Expand ( I(k) ) Around ( k = 0 )

For small ( k ), we can use the Taylor series expansion of ( \sin(\pi k) ):

[
\sin(\pi k) = \pi k – \frac{(\pi k)^3}{6} + \cdots
]

Therefore,

[
\frac{\pi k}{\sin(\pi k)} \approx 1 + \frac{\pi^2 k^2}{6} + \cdots
]

Step 4: Differentiate ( I(k) ) Twice

Taking the second derivative with respect to ( k ):

[
I”(k) = \frac{d^2}{dk^2} \left(1 + \frac{\pi^2 k^2}{6} + \cdots \right) = \frac{\pi^2}{3}
]

Step 5: Conclusion

Thus,

[
\int_{0}^{\infty} \frac{(\ln x)^2}{(1 + x)^2} , dx = I”(0) = \frac{\pi^2}{3}
]

Final Answer:

Its integral is one third of π squared. In symbols,

∫₀^∞ (ln x)²⁄(1 + x)² dx = π²⁄3