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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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# 6 integrals using contours

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