# Integration using residu theorem

• MaxManus
In summary, the integral \int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx can be solved using the method of residues. By choosing a contour with a half-circle in the first and second quadrant and the x-axis, it is possible to calculate the limit of the integral as R approaches infinity. This yields the identity \int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx = \frac12 \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)(x^2+
MaxManus

## Homework Statement

Solve
$$\int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx$$
works here
http://www.codecogs.com/latex/eqneditor.php

using residues

## The Attempt at a Solution

Can I use residues here?
The contour intergral will be a quarter of a circle in the first quadrant and lines along the positive x- and y- axis of

$$\frac{z^2}{(z+i)(z-i)(x+2i)(z-2i)}$$

But can I use Cauchy's Residue Theorem when I get divide by 0 in the contour? i and 2i is on the contour.

Last edited:
Would it help if I noticed that
$$\int_0^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} \, \mathrm dx = \frac12 \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} \, \mathrm dx$$
?

Thanks, that solved the problem
$$\int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx$$
=
$$0.5*2*\pi*i*(res(i) + res(2i)) = \pi*i*(\frac{-1}{6i} +\frac{1}{3i}) = \frac{\pi}{6}$$

Exactly, well done.

(Of course, you would still have to prove that the identity I stated is actually true, and describe which contour you will choose.

Thanks

The identity is true because f(x) = f(-x) so the integral from 0 to infinity is half the integral of f from -infinity to infinity.

I'm using the contour with a half circle in the first and second quadrant with center in (0,0) and radius i equal infinity and the contour along the x-axis from -infinity to infinity.

Ah, so you mean that you are calculating the limit of $R \to \infty$ of
$$\int_{-R}^R \frac{z^2}{(z^2+1)(z^2+4)} dz + \int_{C_R} \frac{z^2}{(z^2+1)(z^2+4)} dz$$

where $C_R: [0, \pi] \to \mathbf{C}, t \mapsto = R e^{i t}$ is a half-circle of radius R.

And you claim that this is equal to the original integral, that is: the second part, which you added to close the contour, does not contribute.

Provided that you all did write this down, just didn't type it out, and proved the claim, I agree :-)

(Sorry for being so annoying

Yes
$$\int_0^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} \, \mathrm dx = \frac12 \int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} \, \mathrm dx$$ = $$0.5* \int_{C_1} \frac{z^2}{(z^2+1)(z^2+4)} dz - 0.5* \int_{C_2} \frac{z^2}{(z^2+1)(z^2+4)}$$

where C_1 is the closed half circle with radius infinity and C_2 is the open half circle with radius infinity.

The C_1 integral is given by Cauchy's Residue Theorem and the C_2 integral is equal to zero.

Thanks for taking the time to help me

## What is the residu theorem?

The residu theorem, also known as the residue theorem, is a mathematical tool used in complex analysis to evaluate certain types of integrals. It states that the value of a contour integral around a closed curve is equal to 2πi times the sum of the residues of the function inside the curve.

## What is integration using residu theorem used for?

Integration using residu theorem is used to evaluate integrals that cannot be solved using traditional methods, such as the fundamental theorem of calculus. It is especially useful in evaluating integrals involving trigonometric, logarithmic, and exponential functions.

## How is the residu theorem applied in practice?

The residu theorem is applied by first identifying the singularities of the function within the contour, then calculating the residues at each singularity. The sum of these residues is then multiplied by 2πi to obtain the value of the integral.

## What are the limitations of integration using residu theorem?

The residu theorem can only be applied to integrals that have singularities within the contour of integration. It also requires knowledge of complex analysis and the ability to identify and calculate residues.

## Can the residu theorem be used for real-valued integrals?

Yes, the residu theorem can be used for real-valued integrals by extending the real-valued function to a complex function and then applying the theorem. However, this is not always necessary as real-valued integrals can often be solved using other methods.

• Calculus and Beyond Homework Help
Replies
3
Views
853
• Calculus and Beyond Homework Help
Replies
23
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
974
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
245
• Calculus and Beyond Homework Help
Replies
8
Views
151
• Calculus and Beyond Homework Help
Replies
1
Views
600
• Calculus and Beyond Homework Help
Replies
2
Views
855
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
972