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1

I don't see any problems up until maybe the very very last step (where you lost some constants at the very least and I can't tell what you were trying to do or why you think there is a problem). So you were at: $$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} x dx.$$ I'm not sure if you then tried a u-substitution or just found an ...

0

Why is that? Where am I getting wrong? The proper approach is to note that the integrand for the $x$ component is an odd function of $x$. But first, recall that $$\int_{-a}^a f(x)dx = \int_{-a}^0 f(x)dx + \int_{0}^a f(x)dx = \int_a^0 f(-x)dx + \int_{0}^a f(x)dx$$ Now, for an odd function, $$f(-x) = -f(x)$$ thus $$\int_a^0 f(-x)dx + \int_{0}^a ... 0 If yes, how I can do it? Try the Feynman Parameter trick. It allows you to take the product of terms in the denominator and convert it into a sum. Then the sum looks like a standard integral. See, for example, Google or Peskin & Schroder at ~page 250. 3 I think i got it: You have$$I=\int \frac{d^4k}{(2\pi)^4}\frac{1}{k^2-m^2_h}\frac{1}{k^2-m^2_\phi}\frac{1}{k^2} in order to calculate this integral in an easy way we make a wick rotation to put ourselves in 4-D euclidean space i.e. $k_0\rightarrow ik_0$ this means that $d^4k=i d^4k$ and $k^2=k_0^2-\vec{k}^2=-k_0^2-\vec{k}^2$ so $k_0^2+\vec{k}^2=-k^2$ if ...

0

The volume of a $N$-sphere is proportional to $R^N$ then the volume of the spherical shell of thickness $dR$ would be proportional to $R^N-(R-dR)^N=R^N(1-(1-dR/R)^N)$, assuming $dR<<R$ but $N(dR)>>R$ you can approximate $$(1-dR/R)^N\simeq\exp{(-dR/R)^N}=\exp{(-NdR/R)}\simeq0.$$ Then the volume of the shell is ...

3

The exponential part factors into a product of exponentials, the sum splits into a sum of integrals and spherical coordinates just mean $dV_1=4\pi r_1^2\,dr_1$ and analogously for $V_2$. What remains will just be a sum of Gamma integrals, you will only need $\int_0^\infty x^n e^{-x}dx=n!$.

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