1. ## Integration by parts

Hello,
How to integrate this integral by using integration by parts. $\Gamma(n)=\displaystyle\int_0^1(\ln{\frac1x})^{n-1}dx$

Someone has answered it as follows:-$\frac{-x}{n}\large(\ln{\frac1x}\large)^n|_{x=0}^{x=1}+ \displaystyle\int_0^1\frac1n \large(\ln{\frac1x}\large)^n=\frac1n \displaystyle\int_0^1\large(\ln{ \frac1x}\large)^n$

And initial case n=1 $\displaystyle\int_0^1 \large(\ln{\frac1x}\large)^0 dx=1$

So by induction $\displaystyle\int_0^1 \large(\ln{\frac1x}\large)^{n-1} dx=(n-1)!$

Now i didn't understand how the integration by parts was done. If anyone knows it,reply it.

2. ## Re: Integration by parts

Originally Posted by Vinod
Hello,
How to integrate this integral by using integration by parts. $\Gamma(n)=\displaystyle\int_0^1(\ln{\frac1x})^{n-1}dx$

Someone has answered it as follows:$\frac{-x}{n}\large(\ln{\frac1x}\large)^n|_{x=0}^{x=1}+ \displaystyle\int_0^1\frac1n \large(\ln{\frac1x}\large)^n=\frac1n \displaystyle\int_0^1\large(\ln{ \frac1x}\large)^n$

And initial case n=1 $\displaystyle\int_0^1 \large(\ln{\frac1x}\large)^0 dx=1$

So by induction $\displaystyle\int_0^1 \large(\ln{\frac1x}\large)^{n-1} dx=(n-1)!$

Now i didn't understand how the integration by parts was done. If anyone knows it,reply it.
I have deleted the minus sign before $\frac{-x}{n}$

3. ## Re: Integration by parts

$u=\left( \ln \dfrac 1 x \right)^n, dv=dx$

4. ## Re: Integration by parts

Originally Posted by SlipEternal
$u=\left( \ln \dfrac 1 x \right)^n, dv=dx$
Hello,
Let say $I_{n-1}=\displaystyle\int \left(\ln{\frac1x}\right)^{n-1}$

$I_{n-1}= \left( \ln{ \frac1x}\right)^{n-1}\displaystyle\int dx -\displaystyle\int\left( \frac{d \left(\ln{\frac1x}\right)^{n-1}}{dx}\displaystyle\int dx \right) dx$

$I_{n-1}=x\left(\ln{\frac1x}\right)^{n-1}-(n-1)\displaystyle\int \left(\ln{\frac1x}\right)^{n-2} dx$

So $I_{n-1}=x\left(\ln{\frac1x}\right)^{n-1}-(n-1)I_{n-2}$

But the answer posted by other fellow is something different in post#1. You are also saying u=\left( \ln\frac1x \right)^n$. How it is? I didn't follow. 5. ## Re: Integration by parts Originally Posted by Vinod Hello, Let say$I_{n-1}=\displaystyle\int \left(\ln{\frac1x}\right)^{n-1} dxI_{n-1}= \left( \ln{ \frac1x}\right)^{n-1}\displaystyle\int dx -\displaystyle\int\left( \frac{d \left(\ln{\frac1x}\right)^{n-1}}{dx}\displaystyle\int dx \right) dxI_{n-1}=x\left(\ln{\frac1x}\right)^{n-1}-(n-1)\displaystyle\int \left(\ln{\frac1x}\right)^{n-2} dx$So$I_{n-1}=x\left(\ln{\frac1x}\right)^{n-1}-(n-1)I_{n-2}$But the answer posted by other fellow is something different in post#1. You are also saying u=\left( \ln\frac1x \right)^n$. How it is? I didn't follow.
Hello,
Please read $I_{n-1}=\displaystyle\int \left(\ln {\frac1x}\right)^{n-1} dx$.

6. ## Re: Integration by parts

Originally Posted by Vinod
Hello,
Please read $I_{n-1}=\displaystyle\int \left(\ln {\frac1x}\right)^{n-1} dx$.
Hello,
Alternatively, we can also say $I_{n-1}=\frac{x}{n} \left(\ln{\frac1x}\right)^n-\frac1nI_n$

7. ## Re: Integration by parts

I gave you the integration by parts that gives the result in your first post. I did not say it had anything to do with your original problem in your first post. I am not sure I understand what you are saying in any of the following posts. You may be making a mistake with the following differentiation.

If

$$u=\left( \ln \dfrac 1 x \right)^{n-1}$$

Find $du$

8. ## Re: Integration by parts

Originally Posted by SlipEternal
I gave you the integration by parts that gives the result in your first post. I did not say it had anything to do with your original problem in your first post. I am not sure I understand what you are saying in any of the following posts. You may be making a mistake with the following differentiation.

If

$$u=\left( \ln \dfrac 1 x \right)^{n-1}$$

Find $du$
Hello,
$du= \frac{n-1}{x} \left( \ln{\frac1x} \right)^{n-2} dx$

9. ## Re: Integration by parts

Originally Posted by Vinod
Hello,
$du= \frac{n-1}{x}\displaystyle\int \left( \ln{\frac1x} \right)^{n-2} dx$
No. That is not how differentiation works. There should be no integral in $du$.

10. ## Re: Integration by parts

Originally Posted by SlipEternal
No. That is not how differentiation works. There should be no integral in $du$.
Hello, I have edited post#8. Look at that post. Whatever i posted in other posts are reduction formulas

11. ## Re: Integration by parts

Originally Posted by Vinod
Hello, I have edited post#8. Look at that post. Whatever i posted in other posts are reduction formulas
You missed a negative sign.

$$du = -\dfrac{(n-1)}{x}\left( \ln \dfrac 1 x \right)^{n-2}dx$$

So,

$$\int_0^1 \left( \ln \dfrac 1 x \right)^{n-1}dx = \left. x\left( \ln \dfrac 1 x \right)^{n-1}\right]_0^1 + (n-1)\int_0^1\left( \ln \dfrac 1 x \right)^{n-2}dx$$

When you plug in 1, you get $\ln \dfrac 1 x =0$. As $x\to 0$, you get 0 times a number which is still zero. So, that first term is zero. You just reduced the exponent by one.