do you agree that
\lim_{x\to+\infty}\;\tan^{-1} x\;=\;\lim_{y\to 0^{+}}\;\tan^{-1} \frac{1}{y}\;=\;\lim_{y\to 0^{+}}\;\left[ \frac{\pi}{2}\;-\;\tan^{-1} y\right]
=\;\frac{\pi}{2}\;-\;\lim_{y\to 0^{+}}\;\sum_{j=0}^\infty\;\frac{(-1)^j\;y^{2j+1}}{2j+1}\;=\;\boxed{\frac{\pi}{2}}\;\;\;\quad...
i was quite surprised to see that \zeta{(5)} crept into the answer. :redface: i am still doing the manual calculation, things are getting quite messy, so i am taking a little break now... but i am still very interested to see how the 5th riemann zeta comes into the picture in both those integrals...
did you actually mean to say since \frac{\mathrm d}{\mathrm dx}\left( \tan^{-1} x\right) \;=\;\frac{1}{1+x^2},
then i could do \textbf u\;=\;\ln {(1+x^2)}\;\;\;\;\;\textbf{and}\;\;\;\;\;\textbf v\;=\;\frac{\left( \tan^{-1} x\right)^3}{1+x^2}\;\;\;?
but then you will get...
hey, thanks. :smile:
[ i don't have mathematica or maple, so i am doing it by hand... i am poor... :frown: ]
anyhow, i will also try and see if i can do
\int_0^\infty\;\;\frac{\ln {(1+x^2)}\cdot \left( \tan^{-1} x\right)^3}{1+x^2}\;\;\mathrm dx
Hello... How can I find an appropriate 'periodic' function (associated with a
Fourier series) to derive the following 2 sums?
1. \displaystyle \sum_{k=1}^{\infty} \,\,\, \frac{\coth{(\pi k)}}{k^{3}} \,\,\, = \,\,\, \frac{7 \pi^{3}}{180}
2. \displaystyle \sum_{k=1}^{\infty} \,\,\...
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...
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...