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The integral of 1 / (1 + x^4) over the real line is pi over sqrt(2)

Example

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

Facts & Assumptions

Given: The rational function $R (z) = 1 / (1 + z^{4})$ .

[L1]

The rational residue theorem evaluates the real integral by the upper-half-plane residues (Rational improper integrals without real poles are upper-half-plane residue sums).

Verification

technique · computation

1.1

The poles of $R$ in the upper half-plane are $a_{1} = e^{i π / 4}$ and $a_{2} = e^{3 i π / 4}$ , and they are simple. Since $(z^{4} + 1)^{'} = 4 z^{3}$ , $Res (R, a_{j}) = \frac{1}{4 a_{j}^{3}} (j = 1, 2) .$

givenalgebra

2.1

Their sum is $\frac{1}{4} (e^{- 3 i π / 4} + e^{- 9 i π / 4}) = \frac{1}{4} (e^{- 3 i π / 4} + e^{- i π / 4}) = - \frac{i}{2 \sqrt{2}} .$

step 1.1algebra

3.1

Therefore [L1] gives $\int_{- \infty}^{\infty} \frac{d x}{1 + x^{4}} = 2 π i (- \frac{i}{2 \sqrt{2}}) = \frac{π}{\sqrt{2}} .$

step 2.1L1∎

Depends on

Rational improper integrals without real poles are upper-half-plane residue sums

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

R. Howell and J. Mathews, Complex Analysis, Ch. 8 §8.3 (standard reference, not scraped)