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bincy
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Dear All,\(\displaystyle \int_{0}^{1}x*\left(\left(-ln(x)\right)^{k}\right)*\left(1-x\right)^{(N-1)}dx \), where k is an odd no. N >=2.regards,
Bincy
Bincy
Hi bincybn,bincybn said:Dear All,\(\displaystyle \int_{0}^{1}x*\left(\left(-ln(x)\right)^{k}\right)*\left(1-x\right)^{(N-1)}dx \), where k is an odd no. N >=2.
regards,
Bincy
Sudharaka said:Hi bincybn,
Your integral cannot be expressed in terms of elementary functions. However using Maxima I found that the answer is,
\[\displaystyle \int_{0}^{1}x((-ln(x))^{k}\left(1-x\right)^{(N-1)}dx=(-1)^k\left( \left. \frac{{d}^{k}}{d\,{x}^{k}}\,\beta\left(N,x\right) \right|_{x=2}\right)~~~~\mbox{for }k\in\mathbb{Z^+}\]
\(\beta\) is the Beta function.
dwsmith said:Since the integral is set up like the beta function, maybe he was supposed to obtain the solution by induction.
Amer said:[tex]\int_{0}^{\infty} e^{-at} t^{z-1} dt = \frac{\Gamma (z)}{a} [/tex]
A definite integral is a mathematical concept used to find the area under a curve between two specified points. It is represented by the symbol ∫ and is calculated by adding up infinitely small rectangles under the curve.
x(-ln(x)^k)(1-x)^(N-1) is a mathematical function that can be used to model various real-life situations. It consists of three parts: x, -ln(x)^k, and (1-x)^(N-1). The value of k and N determines the shape of the curve.
x(-ln(x)^k)(1-x)^(N-1) is an integrand that can be integrated using the definite integral. When the definite integral is applied to this function, it calculates the area under the curve between two specified points.
x(-ln(x)^k)(1-x)^(N-1) has various applications in fields such as physics, economics, and engineering. It can be used to model growth and decay processes, population growth, and financial investments.
To calculate the definite integral of x(-ln(x)^k)(1-x)^(N-1), you can use integration techniques such as substitution, integration by parts, or partial fractions. You can also use online calculators or software to find the numerical value of the integral.