# Thread: Integration by parts using known limit

1. ## Integration by parts using known limit

I have a sum $\displaystyle{ \sum_{k=0}^\infty \frac{1}{k!}\left\{\int_0^1 (x\ln x)^k dx\right\} }$.

Now, per partes it "is easy to show that" $\displaystyle{ \int_0^1 (x\ln x)^k dx=\frac{(-1)^kk!}{(k+1)^{k+1}} }$ (for this I should use $\displaystyle{ \lim_{x\to 0}x\ln x=0 }$).

I "know" that the $\lim$ holds ... but how to use it in per partes integration?

2. the limit does not hold, it's obviously false, can't you just see it?

as for the integral put $t=-\ln x$ and invoke the gamma function.

3. Originally Posted by courteous
I have a sum $\displaystyle{ \sum_{k=0}^\infty \frac{1}{k!}\left\{\int_0^1 (x\ln x)^k dx\right\} }$.

Now, per partes it "is easy to show that" $\displaystyle{ \int_0^1 (x\ln x)^k dx=\frac{(-1)^kk!}{(k+1)^{k+1}} }$ (for this I should use $\displaystyle{ \lim_{x\to \infty}x\ln x=0 }$).

I "know" that the $\lim$ holds ... but how to use it in per partes integration?
Let consider first the indefinite integral...

$\displaystyle \int t^{m}\ \ln ^{n} t\ dt$ (1)

... with m and n non negative integers. Integrating by parts we obtain...

$\displaystyle \int t^{m}\ \ln ^{n} t\ dt = \frac{t^{m+1}}{m+1}\ \ln^{n} t - \frac{n}{m+1}\ \int t^{m}\ \ln^{n-1} t\ dt$ (2)

The integral in (2) is similar to the integral (1), but the exponent of log is n-1 instead of n. That suggests to integrate n times by part so that we obtain...

$\displaystyle \int t^{m}\ \ln ^{n} t\ dt = t^{m+1}\ \sum_{i=0}^{n} \frac{(-1)^{i}}{(m+1)^{i+1}}\ \frac{n!}{(n-i)!}\ \ln ^{n-i} t +c$ (3)

The second step is the computation of...

$\displaystyle \lim_{t \rightarrow 0} t^{m}\ \ln^{n} t$ (4)

... and that is done applying n times l'Hopital's rule...

$\displaystyle \lim_{t \rightarrow 0} t^{m}\ \ln^{n} t = \lim_{t \rightarrow 0} (-1)^{n} \frac{n!}{m^{n}}\ t^{m} = 0$ for $m>0$ (5)

Combining (3) and (5) with some easy steps we arrive to write...

$\displaystyle \int_{0}^{1}t^{m}\ \ln^{n} t\ dt = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}}$ (6)

... that for m=n=k becomes...

$\displaystyle \int_{0}^{1}t^{k}\ \ln^{k} t\ dt = (-1)^{k}\ \frac{k!}{(k+1)^{k+1}}$ (7)

Kind regards

$\chi$ $\sigma$