4

I am being asked the following:

For $n \gt 0$, and by considering an appropriate limit, find:

$$\int_0^1x^n\log(x)\,dx$$

I want to try integration by parts, but I am confused about what the question means when it states by considering an appropriate limit.

Quanto
  • 120,125
DavGRoz
  • 177
  • 6

5 Answers5

8

Note $\int_0^1x^n\,dx= \frac1{n+1}$ and

$$ \int_0^1x^n\log xdx= \frac {d}{dn } \int_0^1x^n\,dx =-\frac1{(n+1)^2}$$

Quanto
  • 120,125
  • 1
    It's worth mentioning that this is Leibniz Rule, more informally known as Feynman's trick. – Deepak Aug 15 '20 at 05:14
2

When approaching limits using integration by parts, you need to ask yourself: "which parts of this integral will become simpler when I integrate or differentiate it?" So let's have a look at $$\int x^n \ln(x)\mathrm{d}x$$ Well, $x^n$ doesn't become much more complicated when you differentiate or integrate it. However, while $\ln(x)$ is difficult to integrate, its derivative is quite simple, namely, $1/x$. So we choose $u=\ln(x)$, $\mathrm{d}v=x^n\mathrm{d}x$, and from this we conclude $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$, $v=\frac{x^{n+1}}{n+1}$. So, $$\int x^n \ln(x)\mathrm{d}x=\int u\mathrm{d}v=uv-\int v\mathrm{d}u=\ln(x)\frac{x^{n+1}}{n+1}-\int \frac{x^{n+1}}{n+1} \frac{1}{x}\mathrm{d}x$$ $$=x^{n+1}\left(\frac{\ln(x)}{n+1}-\frac{1}{(n+1)^2}\right)+C$$ In the limiting case of $n\to 0$ this gives the nice expression $$\int \ln(x)\mathrm{d}x=x\ln(x)-x+C$$

EDIT: This also shows that $$\int_0^1 x^n \ln(x)\mathrm{d}x=1^{n+1}\left(\frac{\ln(1)}{n+1}-\frac{1}{(n+1)^2}\right)=\frac{-1}{(n+1)^2}.$$

K.defaoite
  • 13,890
1

$$\int_0^1x^n\log(x)dx=\left[\log(x)\frac{x^{n+1}}{n+1}\right]_0^1-\int_0^1\frac{x^n}{n+1}dx=-\left[\frac{x^{n+1}}{(n+1)^2}\right]_0^1=-\frac{1}{(n+1)^2}$$

Bachamohamed
  • 1,172
  • Could you please clarify how you eliminated log(x)? – DavGRoz Aug 12 '20 at 22:48
  • 1
    $u(x)=\log(x),\text{and}\quad v^{'}(x)=x^n$ so by using integration by parte we have $\int_a^b u(x)v^{'}(x)dx=(u(x)v(x))_a^b-\int_0^1u^{'}(x)v(x)dx$ $\int_0^1x^n\log(x)dx=\left[\log(x)\frac{x^{n+1}}{n+1}\right]_0^1-\int_0^1(\log(x))^{'}\frac{x^{n+1}}{n+1}dx=-\int_0^1(\frac{1}{x})\frac{x^{n+1}}{n+1}dx=-\int_0^1\frac{x^n}{n+1}dx$ – Bachamohamed Aug 12 '20 at 22:53
  • 1
    @DavGRozNow it looks clear – Bachamohamed Aug 12 '20 at 23:04
  • Yes, a little clearer thank you but I don't understand how to eliminate this term: $[log(x)\frac{x^n+1}{n+1}]_0^1$ because $log(x)$ is undefined for $x = 0$, so would I take the limit as x tends to negative infinity? – DavGRoz Aug 12 '20 at 23:13
  • 1
    $\left[\log(x)\frac{x^{n+1}}{n+1}\right]0^1=\left[\log(1)\frac{1}{n+1}-\log(0)\frac{0}{n+1}\right]=\left[0-\lim{x\to 0}\frac{x^{n+1}}{n+1}\log(x)\right]=\left[0-0\right]=0$ – Bachamohamed Aug 12 '20 at 23:19
  • 1
    Thank you, makes much more sense now! I took the limit as x tends to 0 and got the same solution now. – DavGRoz Aug 12 '20 at 23:22
  • @DavGRoz you welcom. – Bachamohamed Aug 12 '20 at 23:27
1

Using integration by parts:

Let $\displaystyle u=\ln(x) \Leftrightarrow u'=\frac{1}{x}$ and $\displaystyle v=\frac{x^{n+1}}{n+1}\Leftrightarrow v'=x^n$ \begin{align} \mathcal{I}&=\int_0^1 x^n\ln(x)\mathrm{d}x\\ &=\frac{x^n\ln(x)}{n+1}\bigg\vert_0^1-\int_0^1\frac{x^{n+1}}{(n+1)x}\mathrm{d}x\\ &=-\int_0^1\frac{x^n}{n+1}\mathrm{d}x\\ &=-\frac{1}{n+1}\bigg(\frac{x^{n+1}}{n+1}\bigg\vert_0^1\bigg)\\ &=-\frac{1}{(n+1)^2} \end{align} Hence our final answer is: $$\mathcal{I}=-\frac{1}{(n+1)^2}$$ Good luck!

euler_med
  • 1,064
0

Let $x=e^u$ and $dx=e^udu$

Then,

$$\int_{0}^{1} {x^nlnxdx}=\int_{-\infty}^{0} {ue^{un}e^udu}$$ $$=\int_{-\infty}^{0}ue^{(n+1)u}du$$ $$=\bigg[{{(n+1)u-1}\over{(n+1)^2}} {e^{(n+1)u}} ] _{-\infty}^{0}$$ $$=-{1\over {(n+1)^2}}$$

Venkat Amith
  • 475
  • 2
  • 12