- #1
nasim
- 9
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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) \int_{0}^{\infty} \frac{\ln(1+x)}{1+x^{2}} dx
PS: I already know how to solve
\int_{0}^{\infty} \frac{\ln(x)}{1+x^{2}} dx
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 \infty, which
throws me off.
(2) \int_{0}^{\infty} \frac{\ln(1+x+x^{2})}{1+x^{2}} dx
PS: Again, I know how to solve
\int_{0}^{\infty} \frac{\ln(1+x^{2})}{1+x^{2}} dx
which equals \pi \ln 2, but the presence of "x"
within 1+x+x^{2} in (2) is giving me a hard time.
(3) \int_{0}^{\infty} \frac{\ln^{3}(1+x^{2})}{1+x^{2}} dx i.e. \int_{0}^{\infty} \frac{(\ln(1+x^{2}))^{3}}{1+x^{2}} dx
(4) \int_{0}^{\infty} \frac{x \ln(1+x^{2})}{1+x^{2}} dx
(5) \int_{0}^{\infty} \frac{\ln(1+x^{2})}{(1+x^{2}) \sqrt{4+x^{2}}} dx
(6) \int_{0}^{\infty} \frac{\sqrt{x} \ln{(1+x)}}{1+x^{2}} dx
Here, the numerator consists of product of 2 multi-valued functions
with differing branch points within the complex domain
[one at z=0 for \sqrt{z}, 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) \int_{0}^{\infty} \frac{\sqrt{x} \sin^{-1}(1+x)}{1+x^{2}} dx
Again, the numerator here consists of product of 2 multi-valued functions
(sqrt and arc sin); with one branch point at z=0 for \sqrt{z},
and then 2 branch cuts for \sin^{-1}(1+z)
(i) from -\infty to -2, and (ii) from 0 to \infty.
Thanks,
---Nasim (nasim09021975@gmail.com)
I'm trying to solve these 6 improper integrals using calculus of residues.
OK, I have actually got 7 now...
(1) \int_{0}^{\infty} \frac{\ln(1+x)}{1+x^{2}} dx
PS: I already know how to solve
\int_{0}^{\infty} \frac{\ln(x)}{1+x^{2}} dx
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 \infty, which
throws me off.
(2) \int_{0}^{\infty} \frac{\ln(1+x+x^{2})}{1+x^{2}} dx
PS: Again, I know how to solve
\int_{0}^{\infty} \frac{\ln(1+x^{2})}{1+x^{2}} dx
which equals \pi \ln 2, but the presence of "x"
within 1+x+x^{2} in (2) is giving me a hard time.
(3) \int_{0}^{\infty} \frac{\ln^{3}(1+x^{2})}{1+x^{2}} dx i.e. \int_{0}^{\infty} \frac{(\ln(1+x^{2}))^{3}}{1+x^{2}} dx
(4) \int_{0}^{\infty} \frac{x \ln(1+x^{2})}{1+x^{2}} dx
(5) \int_{0}^{\infty} \frac{\ln(1+x^{2})}{(1+x^{2}) \sqrt{4+x^{2}}} dx
(6) \int_{0}^{\infty} \frac{\sqrt{x} \ln{(1+x)}}{1+x^{2}} dx
Here, the numerator consists of product of 2 multi-valued functions
with differing branch points within the complex domain
[one at z=0 for \sqrt{z}, 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) \int_{0}^{\infty} \frac{\sqrt{x} \sin^{-1}(1+x)}{1+x^{2}} dx
Again, the numerator here consists of product of 2 multi-valued functions
(sqrt and arc sin); with one branch point at z=0 for \sqrt{z},
and then 2 branch cuts for \sin^{-1}(1+z)
(i) from -\infty to -2, and (ii) from 0 to \infty.
Thanks,
---Nasim (nasim09021975@gmail.com)
Last edited:
will you please let me know (to 
