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Question: How do you use integration by parts to establish the reduction formula \({\int {\left( {\ln \left...

How do you use integration by parts to establish the reduction formula
(ln(x))ndx=x(ln(x))nn(ln(x))n1dx{\int {\left( {\ln \left( x \right)} \right)} ^n}dx = x{\left( {\ln \left( x \right)} \right)^n} - n{\int {\left( {\ln \left( x \right)} \right)} ^{n - 1}}dx ?

Explanation

Solution

Integration by parts, also known as partial integration, is a method in calculus that calculates the integral of a function product in terms of the integral of their derivative and antiderivative.

Complete step by step answer:
Keep in mind that Integration by parts includes the following:
udv=uvvdu\int {udv = uv - \int {vdu} }
To do so, we all need to come up with a value for uu and another value for dvdv . We have to use the ILATE method for Integration by Parts to find out which terms would fit best.
ILATE method follows this order: Inverse, Logarithm, Algebraic, Trigonometry and Exponential.
This method will assist us in determining which word should be our uu and which should be our dvdv. Our uu should be whichever term in our equation is higher on this chart. We should skip the nn and treat the (lnx)n{\left( {\ln x} \right)^n} on its own in this equation, making (lnx)n{\left( {\ln x} \right)^n} our uu and dxdx our dvdv .
Now we have to differentiate uu and integrate dvdv , hence we get,
ddx(lnx)n=n(lnx)n1(1x)\dfrac{d}{{dx}}{\left( {\ln x} \right)^n} = n{\left( {\ln x} \right)^{n - 1}}\left( {\dfrac{1}{x}} \right) and
dx=x\int {dx = x}
By using integration by parts formula, we get,
(lnx)ndx=x(lnx)nn(lnx)n1(1x)xdx{\int {\left( {\ln x} \right)} ^n}dx = x{\left( {\ln x} \right)^n} - n{\int {\left( {\ln x} \right)} ^{n - 1}}\left( {\dfrac{1}{x}} \right)xdx
Now, xx and 1x\dfrac{1}{x} cancel each other out.
So the final equation becomes,
(lnx)ndx=x(lnx)nn(lnx)n1dx{\int {\left( {\ln x} \right)} ^n}dx = x{\left( {\ln x} \right)^n} - n{\int {\left( {\ln x} \right)} ^{n - 1}}dx

Note: Integration by parts is frequently used in harmonic analysis, particularly Fourier analysis, to demonstrate that continuously oscillating integrals with smooth integrands degrade rapidly. The most common use is to show how the smoothness of a parameter Fourier transform influences its decay.
The research of how general functions can be interpreted or approximated by sums of simpler trigonometric functions is known as Fourier analysis.