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5. Show that [² -1 (k+ 1)²¹ where k> -1 using the gamma function. Hint: consider the substitution u-logx. x log x dx =

5. Show that [² -1 (k+ 1)²¹ where k> -1 using the gamma function. Hint: consider the substitution u-logx. x log x dx =

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How to do question 5.
### Integration and Poisson Distribution in Mathematical Analysis **5. Show that** \[ \int_0^1 x^k \log x \, dx = \frac{-1}{(k+1)^2} \] where \( k > -1 \) using the gamma function. Hint: consider the substitution \( u = -\log x \). **6. The Poisson Distribution is given by** \[ p_n = \frac{\mu^n e^{-\mu}}{n!} \] where \( \mu > 0 \) is the mean and \(\sum_{n=0}^{\infty} p_n = 1\). The Stirling approximation for the Gamma function for large \( z \) is \[ \Gamma(z+1) \approx \sqrt{2\pi} \, z^{z+1/2} e^{-z}. \] Use this approximation to show that the Poisson Distribution becomes a Gaussian Distribution when \( \mu \gg 1 \) and \( n \gg 1 \), i.e., that \[ p_n \approx \frac{e^{-(n-\mu)^2/(2\mu)}}{\sqrt{2\pi \mu}}. \]
Transcribed Image Text:### Integration and Poisson Distribution in Mathematical Analysis **5. Show that** \[ \int_0^1 x^k \log x \, dx = \frac{-1}{(k+1)^2} \] where \( k > -1 \) using the gamma function. Hint: consider the substitution \( u = -\log x \). **6. The Poisson Distribution is given by** \[ p_n = \frac{\mu^n e^{-\mu}}{n!} \] where \( \mu > 0 \) is the mean and \(\sum_{n=0}^{\infty} p_n = 1\). The Stirling approximation for the Gamma function for large \( z \) is \[ \Gamma(z+1) \approx \sqrt{2\pi} \, z^{z+1/2} e^{-z}. \] Use this approximation to show that the Poisson Distribution becomes a Gaussian Distribution when \( \mu \gg 1 \) and \( n \gg 1 \), i.e., that \[ p_n \approx \frac{e^{-(n-\mu)^2/(2\mu)}}{\sqrt{2\pi \mu}}. \]
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Given

using gamma function show that

integral subscript 0 superscript 1 x to the power of k log x d x space equals fraction numerator negative 1 over denominator open parentheses k plus 1 close parentheses squared end fraction

where k>-1


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