How do I solve 6 complex integrals using contours?

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In summary: I have no idea if there is a way to directly calculate these answers.In summary, the conversation is about trying to solve 6 improper integrals using calculus of residues. The first one has a branch point at z=-1 and the person is unsure which contour to use. The second one has a "x" within the function, making it harder to solve. The third one has a triple power of ln(1+x^2). The fourth one has a product of two multi-valued functions with different branch points. The fifth one has a square root and a multivalued function. The sixth one also has a product of two multi-valued functions with different branch points. The seventh one has two branch cuts for the multival
  • #1
nasim
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Hello ppl,


I'm trying to solve these 6 improper integrals using calculus of residues.
OK, I have actually got 7 now...


(1) [tex]\int_{0}^{\infty} \frac{\ln(1+x)}{1+x^{2}} dx[/tex]
PS: I already know how to solve
[tex]\int_{0}^{\infty} \frac{\ln(x)}{1+x^{2}} dx[/tex]
which equals 0, where ln(z) is a multiple-valued function
in the complex domain with branch point z=0.
But I didn't know what contour to use for (1) since
the branch point of ln(1+z) is at z=-1. If I indent it
at z=-1 and use a similar shaped contour to that of
ln(z), I get the contribution from -1 to 0 in addition to
the contribution from 0 to [tex]\infty[/tex], which
throws me off.


(2) [tex]\int_{0}^{\infty} \frac{\ln(1+x+x^{2})}{1+x^{2}} dx[/tex]
PS: Again, I know how to solve
[tex]\int_{0}^{\infty} \frac{\ln(1+x^{2})}{1+x^{2}} dx[/tex]
which equals [tex]\pi \ln 2[/tex], but the presence of "x"
within [tex]1+x+x^{2}[/tex] in (2) is giving me a hard time.


(3) [tex]\int_{0}^{\infty} \frac{\ln^{3}(1+x^{2})}{1+x^{2}} dx[/tex] i.e. [tex]\int_{0}^{\infty} \frac{(\ln(1+x^{2}))^{3}}{1+x^{2}} dx[/tex]


(4) [tex]\int_{0}^{\infty} \frac{x \ln(1+x^{2})}{1+x^{2}} dx[/tex]


(5) [tex]\int_{0}^{\infty} \frac{\ln(1+x^{2})}{(1+x^{2}) \sqrt{4+x^{2}}} dx[/tex]


(6) [tex]\int_{0}^{\infty} \frac{\sqrt{x} \ln{(1+x)}}{1+x^{2}} dx[/tex]
Here, the numerator consists of product of 2 multi-valued functions
with differing branch points within the complex domain
[one at z=0 for [tex]\sqrt{z}[/tex], the other at z=-1 for ln(1+z)].
How do I tackle 2 branch points and what would be the best
contour to use here?


(7) [tex]\int_{0}^{\infty} \frac{\sqrt{x} \sin^{-1}(1+x)}{1+x^{2}} dx[/tex]
Again, the numerator here consists of product of 2 multi-valued functions
(sqrt and arc sin); with one branch point at z=0 for [tex]\sqrt{z}[/tex],
and then 2 branch cuts for [tex]\sin^{-1}(1+z)[/tex]
(i) from [tex]-\infty[/tex] to -2, and (ii) from 0 to [tex]\infty[/tex].


Thanks,
---Nasim (nasim09021975@gmail.com)
 
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  • #2
Were you given these for homework or just wondered how to do them? Because the solutions I get for some of these are extremely difficult.

My solution to number 2, which agrees with Mathematica, is 8 lines long and contains non elementary functions.
 
  • #3
Gib Z said:
Were you given these for homework or just wondered how to do them? Because the solutions I get for some of these are extremely difficult.

My solution to number 2, which agrees with Mathematica, is 8 lines long and contains non elementary functions.

wow, that is cool, Gib Z ! :eek: will you please let me know (to
the best of your ability) how you solved them, at least if I can
observe/witness/critic your way of thinking, I believe I can
learn more of the amazing and beautiful topic of residue calculus!

after all, I am here to learn, aren't I ? o:)
 
  • #4
Actually I didn't do them by hand, I probably couldn't. I am just telling you that a Computer Program that is very intelligent gets a very long answer.
 
  • #5
Mathematica often gives complicated answers with stuff like the LambertW function even if the answer can be simple.
 

1. What are integrals using contours?

Integrals using contours involve using the concept of a contour, which is a closed path on a graph, to evaluate integrals. This method is particularly useful for complex integrals.

2. How do you solve integrals using contours?

To solve integrals using contours, you need to follow a few steps. First, you need to choose a contour that encloses the region of interest. Then, you need to parametrize the contour and express the integral in terms of the parametrization. Finally, you can use techniques such as Cauchy's integral formula or the residue theorem to evaluate the integral.

3. What is the difference between closed and open contours?

A closed contour is a path that starts and ends at the same point, forming a loop. An open contour, on the other hand, does not form a loop and has distinct starting and ending points. Closed contours are typically used in integrals using contours as they allow for the use of techniques such as the residue theorem.

4. Can integrals using contours be used for any type of integral?

No, integrals using contours are typically used for integrals involving complex functions. They are not suitable for all types of integrals, such as those involving discontinuous or unbounded functions.

5. What are some applications of integrals using contours?

Integrals using contours have many practical applications, particularly in physics and engineering. They can be used to solve problems involving electric fields, fluid dynamics, and heat transfer, among others. They are also commonly used in complex analysis and the study of special functions.

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