Difference between revisions of "Integral of (z^n)log(z)dz=(z^(n+1)/(n+1))log(z)-z^(n+1)/(n+1)^2 for integer n neq -1"
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(Created page with "==Theorem== The following formula holds for integers $n \neq -1$: $$\displaystyle\int z^n \log(z) \mathrm{d}z= \dfrac{z^{n+1}\log(z)}{n+1} - \dfrac{z^{n+1}}{(n+1)^2},$$ where...") |
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==Theorem== | ==Theorem== | ||
The following formula holds for integers $n \neq -1$: | The following formula holds for integers $n \neq -1$: | ||
| − | $$\displaystyle\int z^n \log(z) \mathrm{d}z= \dfrac{z^{n+1}\log(z)}{n+1} - \dfrac{z^{n+1}}{(n+1)^2},$$ | + | $$\displaystyle\int z^n \log(z) \mathrm{d}z= \dfrac{z^{n+1}\log(z)}{n+1} - \dfrac{z^{n+1}}{(n+1)^2}+C,$$ |
where $\log$ denotes the [[logarithm]]. | where $\log$ denotes the [[logarithm]]. | ||
Latest revision as of 02:01, 24 February 2018
Theorem
The following formula holds for integers $n \neq -1$: $$\displaystyle\int z^n \log(z) \mathrm{d}z= \dfrac{z^{n+1}\log(z)}{n+1} - \dfrac{z^{n+1}}{(n+1)^2}+C,$$ where $\log$ denotes the logarithm.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.1.50$
