Expert Answer to "Finding:∫−∞∞x2x4+1"

daurialebraslmc

Answered question

2022-04-06

Finding:

\int_{- \infty}^{\infty} \frac{x^{2}}{x^{4} + 1}

Assorrymarf0cgr

Beginner2022-04-07Added 10 answers

Notice that:

\int_{- \infty}^{+ \infty} \frac{x^{2}}{x^{4} + 1} d x = 2 \int_{0}^{+ \infty} \frac{x^{2}}{x^{4} + 1} d x

= 2 \int_{0}^{1} \frac{x^{2}}{1 + x^{4}} d x + 2 \int_{1}^{+ \infty} \frac{x^{2}}{1 + x^{4}} d x

= 2 \int_{0}^{1} \frac{1 + x^{2}}{1 + x^{4}} d x

= 2 (1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \dots)

= 2 (1 + \frac{2}{4^{2} - 1} - \frac{2}{8^{2} - 1} + \frac{2}{12^{2} - 1} - \dots)

and, from the logarithmic derivative of the Weierstrass product for the sine and cosine function:

\sum_{k \geq 0} \frac{1}{{(2 k + 1)}^{2} - x^{2}} = \frac{π}{4 x} \tan (\frac{π x}{2})

\sum_{k \geq 1} \frac{1}{k^{2} - x^{2}} = \frac{1 - π x \cot (π x)}{2 x^{2}}

,
so by taking limits as

x \to \frac{1}{2}

we get:

\int_{- \infty}^{+ \infty} \frac{x^{2}}{x^{4} + 1} d x = \frac{π}{\sqrt{2}}

blessgansgxei

Beginner2022-04-08Added 10 answers

Notice

\int_{- \infty}^{\infty} \frac{x^{2}}{x^{4} + 1} d x

= 2 \int_{0}^{\infty} \frac{x^{2}}{x^{4} + 1} d x

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

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

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

Change variable from x to

\frac{1}{x}

for that part of integral on

(0, 1)

, this becomes

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

Change variable once more to

y = x - x^{- 1}

and then

y = \sqrt{2} z

, we get

\int_{- \infty}^{\infty} \frac{x^{2}}{x^{4} + 1} d x = 2 \int_{0}^{\infty} \frac{d y}{y^{2} + 2}

= \int_{- \infty}^{\infty} \frac{d y}{y^{2} + 2}

= \frac{1}{\sqrt{2}} \int_{- \infty}^{\infty} \frac{d z}{1 + z^{2}}

= \frac{π}{\sqrt{2}}

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