$\displaystyle \[\sqrt {{b^2} - 4ac} \]$
am quite new to this forum and i have little idea on how to use the editor but please this is my question
Please am stuck with the integration I kindly need assistance in solving this equation, I will really appreciate if anyone could help with the solution or a guide on what i should do
[tex]C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz[\tex]
The closing tag should be [/TEX], not [\TEX].
$\displaystyle C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz$
Also it is better to write \log rather than log to indicate that it is an operator name rather than the product of three variables $l,o,g$.
$\displaystyle C(N)=\int_{0}^{\infty}\log(1+\frac{\rho}{a}z)\frac { N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz$
Actually I just noticed something else that would make it look better: Code the log as \ln, not simply ln. So now we have either $\int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$ or
$\displaystyle \int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$,
depending on whether it's inline or not.