- #1
skrat
- 748
- 8
Homework Statement
After successfully solving a lot of integrals I gathered 4 ugly ones that I can not solve:
a) ## \int _{-\infty} ^\infty \frac{cos(2x)}{x^4+1}dx##
b) ##\int _0 ^\infty \frac{dx}{1+x^3}##
c) ##\int _0 ^\infty \frac{x^2+1}{x^4+1}dx##
d) ##\int _0 ^{2\pi } \frac{d\varphi }{a+cos(\varphi )}##
Homework Equations
The Attempt at a Solution
I simply don't know what to do with integrals ##\int _0 ^\infty##. Have no idea! Please help!
I can only show my work and a) and d):
a)
##\int_{-\infty }^{\infty }\frac{cos(2z)}{z^4+1}dz=\int_{-R }^{R }\frac{cos(2x)}{x^4+1}dx+\int _\gamma\frac{cos(2z)}{z^4+1}dz## where ##R->\infty##.
Last integral is for ##z=Re^{i\varphi}## where ##R->\infty## clearely ##0##, therefore only one integral still remains:
##\int_{-\infty }^{\infty }\frac{cos(2z)}{z^4+1}dz##
##\int_{-\infty }^{\infty }\frac{cos(2z)}{z^4+1}dz=2\pi i \sum Res(f,a)##
##z^4+1=0## gives me ##z_1=e^{i\pi /4}##, ##z_2=e^{i3pi /4}##, ##z_3=e^{i5\pi /4}## and ##z_4=e^{i7\pi /4}## singularity points which are all poles of order 1.
Therefore for all singularity points ##Res_{z=z_i} \frac{f(z)}{g(z)}=\frac{f(z_i)}{g^{'}(z_i)}=\frac{cos(2z_i)}{4z_i^3}##.
Now this is where it all stops for me. How much is for example ##\frac{cos(2e^{i5\pi /4})}{3e^{i15\pi /4}}## ?
d)
##\int _0 ^{2\pi } \frac{d\varphi }{a+cos(\varphi )}##
Hmmm... Can I say that ##cos(\varphi )=\frac{z+\bar{z}}{2}##?
Last edited: