Question

Using residu theorem solve integrale ∫(1/(1+x^4))dx integrale from -∞ to +∞ ?? many thanks

Answer 1

`I = int_(-oo)^(oo) 1/(1+x^4) \ dx = (pisqrt(2))/2`

Explanation:

Sorry that this is such a long solution:

We seek:

`I = int_(-oo)^(oo) 1/(1+x^4) \ dx`

graph{1/(1+x^4) [-5, 5, -2.5, 2.5]}

The function is well defined over the domain `RR`. In order to compute this definite integral, consider the following complex variable function over a domain `CC`:

`f(z) =1/(1+z^4)`

And its associated contour integral:

`oint_C \ f(z) \ dz`

Where `C` is the following semi-circulare contour in the complex plane with radius `R gt 0`:

We will restrict `R` to enclose the poles in the upper quadrants once we have analyzed the poles of `f(z)`. The denominator of the integrand is a quartic, and so we can find the four simple poles of `f(z)` as follows:

`z^4+1 = 0 => z^4=-1`

We can readily solve this equation by putting the equation into polar form and using DeMoivre's Theorem

`z^4 = cospi+isinpi => z = {cos(pi+2npi)+isin(pi+2npi)}^(1/4)`
`:. z = cos((pi+2npi)/4)+isin((pi+2npi)/4)`
`\ \ \ \ \ \ = costheta+isintheta` where `theta=((pi+2npi)/4), n in NN`

`{: (n, theta, z, "designation"), (0, pi/4, sqrt(2)/2+isqrt(2)/2, omega), (1, (3pi)/4, sqrt(2)/2-isqrt(2)/2, omega^3), (1, (5pi)/4, -sqrt(2)/2-isqrt(2)/2, omega^5), (1, (7pi)/4, sqrt(2)/2-isqrt(2)/2, omega^7) :}`

We need only be concerned with the two poles in Q1 and Q2, that (providing we make `R` large enough) lie within our contour `C`

So then:

`oint_C \ f(z) \ dz = int_(-R)^(R) \ f(x) \ dx + int_(gamma_R) \ f(z) \ dz`

If we denote the first solution by `omega` then the remaining roots are `omega^3`, `omega^5` and `omega^7`. We can make `R` sufficiently large that it always encloses the two poles in the upper quadrants. `C` then encloses the two simple poles `omega` and `omega^3`/

Then by the residue theorem:

`oint_C \ f(z) \ dz = 2pii xx ( ("sum of the residues of the"), ("poles of " f(z) " within "C) )`
`" " = 2pii \ { res_(z=omega) \ f(z) + res_(z=omega_3) \ f(z) }`

We have simple poles, so:

`1/(1+z^4) = 1/((z-omega)(z-omega^3)(z-omega^5)(z-omega^7))`

And we can calculate the residues as follows:

`res_(z=omega) = lim_(z rarr omega) (z-omega)/(1+z^4)`
`" " = lim_(z rarr omega) (1)/(4z^3)` (using L'Hôpital's rule)
`" " = 1/4omega^-3`
`" " = 1/4e^((5pi)/4i)`

Similarly:

`res_(z=omega^3) = 1/4e^((7pi)/4i)`

So using these results along with the residue theorem we get:

`oint_C \ f(z) \ dz = (2pi i) { 1/4e^((5pi)/4i) + 1/4e^((7pi)/4i)}`
`" " = (2pi i) { 1/4(-sqrt(2)/2-sqrt(2)/2i) + 1/4(sqrt(2)/2-sqrt(2)/2i)}`
`" " = (2pi i) ((-sqrt(2)i)/4 )`
`" " = (pisqrt(2))/2`

Earlier we established that:

`oint_C \ f(z) \ dz = int_(-R)^(R) \ f(x) \ dx + int_(gamma_R) \ f(z) \ dz`

Now, as we let `R rarr oo` we get:

`oint_C \ f(z) \ dz = int_(-oo)^(oo) \ f(x) \ dx + int_(gamma_R) \ f(z) \ dz`

`:. (pisqrt(2))/2 = int_(-oo)^(oo) 1/(1+x^4) \ dx + int_(gamma_R) \ f(z) \ dz`

As is often the case with contour integrals, we find that:

`lim_(R rarr oo) int_(gamma_R) \ f(z) \ dz = 0`

I will omit the proof (as it is several pages long) but this can be verified using the Estimation Lemma. Hence, in summary we have:

`int_(-oo)^(oo) 1/(1+x^4) \ dx = (pisqrt(2))/2`