Get the solution to "I want to find a nice simple expression..."

Leonard Montes

Answered question

2022-04-08

I want to find a nice simple expression for the definite integral

\int_{0}^{\infty} \frac{x^{2} d x}{{(x^{2} - a^{2})}^{2} + x^{2}}

Jax Burns

Beginner2022-04-09Added 13 answers

Let

I = \int_{0}^{\infty} \frac{x^{2}}{x^{2} + {(x^{2} - a^{2})}^{2}} d x

Make the substitution

x = \frac{a^{2}}{t}

We get

I = \int_{0}^{\infty} \frac{a^{6}}{t^{4} (\frac{a^{2}}{t^{2}} + {(\frac{a^{4}}{t^{2}} - a^{2})}^{2})} d t = \int_{0}^{\infty} \frac{a^{2}}{t^{2} + {(a^{2} - t^{2})}^{2}} d t

i.e.

I = \int_{0}^{\infty} \frac{a^{2}}{x^{2} + {(a^{2} - x^{2})}^{2}} d x

Therefore

2 I = \int_{0}^{\infty} \frac{x^{2}}{x^{2} + {(x^{2} - a^{2})}^{2}} d x + \int_{0}^{\infty} \frac{a^{2}}{x^{2} + {(a^{2} - x^{2})}^{2}} d x

= \int_{0}^{\infty} \frac{x^{2} + a^{2}}{x^{2} + {(x^{2} - a^{2})}^{2}} d x

= \int_{0}^{\infty} \frac{1 + \frac{a^{2}}{x^{2}}}{1 + {(x - \frac{a^{2}}{x})}^{2}} d x

Making the substitution

t = x - \frac{a^{2}}{x}

Gives us

2 I = \int_{- \infty}^{\infty} \frac{d t}{1 + t^{2}} = π

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