How to solve the integral using the residue theorem int_(-oo)^(+oo)(cos(x)/(x^4+x^2+1))dx?

1 Answer
Aug 1, 2017

int_(-oo)^(oo) ( cosx ) / ( x^4 + x^2 +1 ) \ dx = e^(-sqrt(3)/2)pi{ sqrt(3)/3cos(1/2) +sin(1/2) }
" " = 1.303046536141762 ...

Explanation:

Sorry that this is such a long solution:

We seek:

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

graph{( cosx ) / ( x^4 + x^2 +1 ) [-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) = ( e^(iz) ) / ( z^4 + z^2 +1 ) \ \ \ \ \ (= ( cosz+isinz ) / ( z^4 + z^2 +1 ) )

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:

enter image source here

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 quadratic in z^2, and so we can find the four simple poles of f(z) as follows:

z^4+z^2+1 = 0 => =0

z^2-z+1 = 0 => z = \ \ \ \ 1/2+-sqrt(3)/2i
z^2+z+1 = 0 => z = -1/2+-sqrt(3)/2i

enter image source here

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

As C encloses the two poles:

z_1 = -1/2+sqrt(3)/2i \ \ and \ \ z_2 = 1/2+sqrt(3)/2i

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=z_1) \ f(z) + res_(z=z_2) \ f(z) }

And we can calculate the residues as follows:

res_(z=z_1) = lim_(z rarr z_1) (z-z_1) f(z)
" " = lim_(z rarr z_1) (z-z_1) ( e^(iz) ) / ( (z^2-z+1)(z^2+z+1) )
" " = lim_(z rarr z_1) (z-z_1) ( e^(iz) ) / ( (z^2-z+1)(z-z_1)(z-barz_1) )
" " = lim_(z rarr z_1) ( e^(iz) ) / ( (z^2-z+1)(z-barz_1) )

" " = e^(-sqrt(3)/2+1/2i) (-1/4-sqrt(3)/12i)
" " = e^(-sqrt(3)/2)e^(1/2i) (-1/4-sqrt(3)/12i)

Similarly:

res_(z=z_2) = lim_(z rarr z_2) (z-z_2) f(z)
" " = lim_(z rarr z_2) (z-z_2) ( e^(iz) ) / ( (z^2-z+1)(z^2+z+1) )
" " = lim_(z rarr z_2) (z-z_2) ( e^(iz) ) / ( (z-z_2)(z-barz_2)(z^2+z+1) )

" " = lim_(z rarr z_2) ( e^(iz) ) / ( (z-barz_2)(z^2+z+1) )

" " = e^(-sqrt(3)/2-1/2i) (1/4-sqrt(3)/12i)
" " = e^(-sqrt(3)/2)e^(-1/2i) (1/4-sqrt(3)/12i)

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

oint_C \ f(z) \ dz = 2e^(-sqrt(3)/2)pii{ -1/4e^(1/2i)-sqrt(3)/12e^(1/2i)i + 1/4e^(-1/2i)-sqrt(3)/12e^(-1/2i)i }

" " = 2e^(-sqrt(3)/2)pi{ -1/4e^(1/2i)i+sqrt(3)/12e^(1/2i) + 1/4e^(-1/2i)i+sqrt(3)/12e^(-1/2i) }

" " = 2e^(-sqrt(3)/2)pi{ sqrt(3)/12(e^(1/2i) + e^(-1/2i)) +1/4(e^(-1/2i)-e^(1/2i))i }

We can simplify the expression as:

e^(1/2i) + e^(-1/2i) = cos(1/2)+isin(1/2) + cos(-1/2)+isin(-1/2)
" " = cos(1/2)+isin(1/2) +cos(1/2)-isin(1/2)
" " = 2cos(1/2)

And:

e^(-1/2i)-e^(1/2i) = cos(-1/2)+isin(-1/2) -cos(1/2)-isin(1/2)
" " = cos(1/2)-isin(1/2) -cos(1/2)-isin(1/2)
" " = e^(-1/2i)-e^(1/2i) = -2isin(1/2)

Hence:

oint_C \ f(z) \ dz =2e^(-sqrt(3)/2)pi{ sqrt(3)/12(2cos(1/2)) +1/4(2sin(1/2)) }
" " = e^(-sqrt(3)/2)pi{ sqrt(3)/3cos(1/2) +sin(1/2) }

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

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) ( e^(iz) ) / ( z^4 + z^2 +1 ) \ dz = int_(-oo)^(oo) ( cosx +isinx) / ( x^4 + x^2 +1 ) \ dx

" " = int_(-oo)^(oo) ( cosx) / ( x^4 + x^2 +1 ) \ dx + i \ int_(-oo)^(oo) ( sinx) / ( x^4 + x^2 +1 ) \ dx

" " = e^(-sqrt(3)/2)pi{ sqrt(3)/3cos(1/2) +sin(1/2) }

Equating Real parts we then get the desired result:

int_(-oo)^(oo) ( cosx ) / ( x^4 + x^2 +1 ) \ dx = e^(-sqrt(3)/2)pi{ sqrt(3)/3cos(1/2) +sin(1/2) }
" " = 1.303046536141762 ...

As a bonus, by equating imaginary parts, we also get

int_(-oo)^(oo) ( sinx ) / ( x^4 + x^2 +1 ) \ dx = 0

But that should come as no surprise as the integrand is an odd function!