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\begin{document}

{\bf Question}

Find a reduction formula for $I_N=\ds\int (\ln x)^n \,dx$.

Evaluate $\ds\int_1^e (\ln x)^4 \,dx$.


\medskip

{\bf Answer} $I_n=\ds\int(\ln x)^n \,dx$

Integrate by parts with

$$\begin{array} {cc} u=(\ln x)^n & \ds\frac{dv}{dx}=1\\
\ds\frac{du}{dx}=n(\ln x)^{n-1} \times \ds\frac{1}{x} & v=x
\end{array}$$

Therefore $\begin{array} {rcl} I_n & = & x(\ln x)^n-\ds\int \,dx
\ds\frac{n(\ln x)^{n-1}}{x} \times x\\ & = & x(\ln x)^n-n\ds\int
\,dx (\ln x)^{n-1} = x(\ln x)^n-nI_{n-1} \end{array}$

Hence $\un{I_n=x(\ln x)^n-n I_{n-1}}$ is the reduction formula.

If $I_4=\ds\int_1^e(\ln x)^4 \,dx$ then inserting the integration
limits into the above expression we have, with $n=4$.

$$\begin{array} {rcl} I_4 & = & [x(\ln x)^4]_1^e -4I_3\\ I_3 & = &
[x(\ln x)^3]_1^e -3I-2\\ I-2 & = & [x(\ln x)^2]_1^e -2I_1\\ I_1 &
= & [x(\ln x)]_1^e -I_0 \end{array}$$

What's $I_0$?

$I-0=\ds\int_1^e (\ln x)^0 \,dx = \ds\int_1^e \,dx = \un{e-1}$

Therefore \begin{eqnarray*} I_1 & = & (e \ln e -1 \ln 1)-(e-1)\ \
\ (\ln 1=0,\ \ln e=1)\\ & = & e-(e-1)=\un{1}\\ \rm{Hence}\ I_2 & =
& [e(\ln e)^2-1(\ln 1)^2]-2I_1\\ & = & e-2\times 1=\un{e-2}\\
\rm{Hence}\ I_3 & = & [e(\ln e)^3 -1(\ln 1)^3]-3I_2\\ & = &
e-3\times (e-2)=\un{6-2e}\\ \rm{Hence}\ I_4 & = & [e(\ln e)^4
-1(\ln 1)^4]-4I_3\\ & = & e-4\times (6-2e)=\un{9e-24}
\end{eqnarray*}
\end{document}
