How to derive a reduction formula for $\displaystyle\int(\ln x)^m\, dx$ where $m$ is a natural number?

Question

How to derive a reduction formula for $\int(\ln x)^m\, dx$ where $m$ is a natural number?

I took $(\ln x)^m$ as the first function and $1$ as the second function.

I am getting the final result $x (\ln x)^m -mx\ln x +mx + C$. But it does not satisfies, for example, $\int(lnx)^4 \,dx$

Where did I go wrong? Please derive the formula.

Answer 1

Well, we have that:

$$\mathscr{I}_\text{n}:=\int\ln^\text{n}\left(x\right)\space\text{d}x\tag1$$

Using integration by parts:

$$\int\text{f}\left(x\right)\cdot\text{g}\space'\left(x\right)\space\text{d}x=\text{f}\left(x\right)\cdot\text{g}\space\left(x\right)-\int\text{f}\space'\left(x\right)\cdot\text{g}\left(x\right)\space\text{d}x\tag2$$

So, we set:

• $$\text{f}\left(x\right)=\ln^\text{n}\left(x\right)\space\space\space\Longleftrightarrow\space\space\space\text{f}\space'\left(x\right)=\frac{\text{n}}{x}\cdot\ln^{\text{n}-1}\left(x\right)\tag3$$
• $$\text{g}\space'\left(x\right)=1\space\space\space\Longleftrightarrow\space\space\space\text{g}\left(x\right)=x\tag4$$

So, we get:

$$\mathscr{I}_\text{n}:=\int\ln^\text{n}\left(x\right)\space\text{d}x=x\ln^\text{n}\left(x\right)-\int\frac{\text{n}}{x}\cdot\ln^{\text{n}-1}\left(x\right)\cdot x\space\text{d}x=$$ $$x\ln^\text{n}\left(x\right)-\text{n}\int\ln^{\text{n}-1}\left(x\right)\space\text{d}x=x\ln^\text{n}\left(x\right)-\text{n}\cdot\mathscr{I}_{\text{n}-1}\tag5$$

So, we get:

$$\mathscr{I}_\text{n}=x\ln^\text{n}\left(x\right)-\text{n}\cdot\mathscr{I}_{\text{n}-1}\space\Longleftrightarrow\space\mathscr{I}_\text{n}+\text{n}\cdot\mathscr{I}_{\text{n}-1}=x\ln^\text{n}\left(x\right)\tag6$$

Answer 2

Let $I_{m} = \int (\ln x)^{m}\, dx$

Let $u = (\ln x)^m\implies du = \frac{m(\ln x)^{m-1}}{x}dx$

Let $v = \int 1 dx = x$

$\implies I_{m} = x(\ln x)^{m} -\int m(\ln x)^{m-1}\, dx = x(\ln x)^m - mI_{n-1}$

Answer 3

Your mistake is in the application of the by-part integration rule and/or the way you differentiated.

You correctly started with

$$\int 1\cdot\log^mx\,dx=x\cdot\log^mx-\text{other integral}$$

Now the new integrand must be

$$x\cdot\left(\log^m x\right)'=x\cdot m\log^{m-1}x\cdot\frac1x.$$

Hence the seeked reduction:

$$I_m=x\cdot\log^mx-mI_{m-1}.$$