An Integral with a Surprising Result!


Good morning fellow mathematicians! Once again it is time for some sweet, sweet mathemagic. In today's post we are going to deal with an integral, which is going to deliver quite the surprising result. Without further ado, let us take a look at what we are going to deal with today:

$$\int_{0}^{1}\frac{\ln{(x)}}{x-1}dx$$

This in itself might not seem to like the most exciting integral in mankind's glorious history, but let us continue and see what we get. At first I would like to introduce a substitution, namely $t=x-1$. But don't forget your differentials! This being said, we are going to arrive at

$$\int_{-1}^{0}\frac{\ln{(t+1)}}{t}dt\text{ ,}$$

where $dt=dx$, with our boundaries changing respectively. Next, we would like to derive the Taylor Series of $\ln{(1+x)}$ using some simple observations.

Consider the derivative of $\ln{(t+1)}$ with respect to $t$. Naturally we are going to get $\frac{1}{1+t}$. This result though, looks suspiciously like the geometric series, just with a sign switched around. Thus, rewriting the fraction as $\frac{1}{1-(-t)}$, we can express it in the  following way:

\begin{align*}
\frac{1}{1-(-t)}&=\sum_{k=0}^{\infty}(-t)^k\\
&=\sum_{k=0}^{\infty}(-1)^kt^k
\end{align*}

Integrating this newly acquired expression is going to provide us with the following result:

$$\int \frac{dt}{1+t}=\int\sum_{k=0}^{\infty}(-1)^kt^kdt$$

Notice that, after doing the integration, we are going to arrive at $\ln{(1+t)}$ once again, where our boundaries could be applied if they would coincide with our geometric series' radius of convergence. Now here comes the difficult part. If our sum were finite, then we could just drag it to the front of the integral and integrate our expression term by term. Sadly, this isn't the case since we are dealing with a series here, meaning us interchanging sum and integral would result in a change of limits. And this is by no means a trivial matter. Let me now try to justify, that we are indeed allowed to do the following:

$$\int\sum_{k=0}^{\infty}(-1)^kt^kdt=\lim\limits_{N \rightarrow\infty}\sum_{k=0}^{N}\int(-1)^kt^kdt$$

Real analysis provides us with a lot of magnificent theorems. Some of the more popular ones which would allow this interchange of limits, are Fubini's/Tonelli's Theorem or the Dominated Convergence Theorem. In our case, the latter on is going to work out just fine! Here's a quick summary of the theorem:


The Dominated Convergence Theorem (DCT):
Let $f,f_1,f_2,...$ and $D$ (the Dominator) be infinite, real valued sequences, such that
$\quad i$)      $\lim\limits_{N\rightarrow\infty}f_N(t)=f(t)$ for all $t$
$\quad ii$)     $|f_N(t)|\le D(t)$ for all $N,t$
$\quad iii$)    $D$ is (riemann) integrable
Then all the sequence members $f_N$ and their limit $f$ are integrable, with:
$\quad\lim\limits_{N\rightarrow\infty}\int f_N=\int f$

Note, that our $f_N$ are in this case the partial sums $\sum_{k=0}^{N}(-1)^kt^k$ and whenever our $|t|<1$, then our members of the partial sum are monotonically decreasing in magnitude. This just means that the first term of our partial sum, namely $1$, can be used as the Dominator $D$. Thus, the $2^{nd}$ condition has been satisfied, with $|\sum_{k=0}^{N}(-1)^kt^k| \le D$. Also, we have our other two conditions being met, since $1$ is just a constant and thus (riemann) integrable and we have our sequence converging pointwise for all $-1<t<1$.

With those technicalities out of the way, let us continue the integration! Overall we now have, that:

$$
\begin{align*}
\int\sum_{k=0}^{\infty}(-1)^k t^k dt&=\sum_{k=0}^{\infty}\int(-1)^kt^kdt\\
&=\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}t^{k+1}\\
&=\ln{(t+1)}
\end{align*}
$$

Now that we have another series representation that we can use, let us plug this one into our original integral and see what we get:

$$
\begin{align*}
\int_{-1}^{0}\frac{\ln{(t+1)}}{t}dt&=\int_{-1}^{0}\frac{1}{t}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}t^{k+1}dt\\
&=\int_{-1}^{0}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}t^{k}dt\\
\end{align*}
$$

And here we go again. Yet another interchange of limits is required in order to solve the underlying integral. But as always, I'm going to leave it as a little exercise for the reader to figure out the details. Once we've switched summation and integration it is a trivial task to integrate the polynomial in $t$, providing us with:

$$
\begin{align*}
\int_{-1}^{0}\frac{\ln{(t+1)}}{t}dt&=\int_{-1}^{0}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}t^{k}dt\\
&=\sum_{k=0}^{\infty}\int_{-1}^{0}\frac{(-1)^k}{k+1}t^{k}dt\\
&=\sum_{k=0}^{\infty}\frac{(-1)^k}{(k+1)^2}t^{k+1}\Biggr|_{-1}^{0}\\
&=0-\sum_{k=0}^{\infty}\frac{(-1)^k}{(k+1)^2}(-1)^{k+1}\\
&=-(-1)\sum_{k=0}^{\infty}\frac{(-1)^{2k} }{(k+1)^2}\\
&=\sum_{k=0}^{\infty}\frac{1}{(k+1)^2}\\
\end{align*}
$$

Our final expression is yet another infinite sum, which might not seem familiar at first glance. But once we introduce a small change of index, namely $k+1=n$, it's going to turn into quite the famous identity:

$$
\begin{align*}
\sum_{k=0}^{\infty}\frac{1}{(k+1)^2}&=\sum_{n=1}^{\infty}\frac{1}{n^2}\\
&=\frac{\pi^2}{6}
\end{align*}
$$

What a great result! Suddenly our most favourite transcendental friend $\pi$ pops up, providing us with a really satisfying answer to the given problem. This last sum is called the Basel Problem by the way. You can either watch my video on it or wait for another article which is going to cover this curious series and its evaluation.

I hope you did enjoy the read and if you did, make sure to subscribe to the blog or activate the push notifications to stay updated! I thank you guys for your attention and up until the next post,

  --have a flammable day $\int d \tau$

Comments

  1. I love it. Thank you Papa Flammy!

    ReplyDelete
  2. A Taylor series expansion always makes up for good integrals. Keep uploading :)

    ReplyDelete
  3. The Taylor series of in(x) is Here defined (-1,1) right? The Question is What if would use the Taylor series which defines the ln(x+1) (-1,Infinity) is there a Problem with the Dominantes convergence? Or can we Just use some different Dominator Thema 1? And If so do WE have any difference?

    ReplyDelete
  4. Typo. "histpry" in the second paragraph.

    ReplyDelete
  5. I like that you included the details about exchanging limits! Pi is always sneaking about.

    Another way to justify it could be that since (-t)^k is intergrable (polynomial), the partial sum of these polynomials is also integrable due to additivity. One can show that this partial sum is uniformly Cauchy and hence uniformly convergent. From this fact we can conclude that the infinite sum of (-t)^k is integrable. Therefore the limit as n tends to infinity of the integral of the partial sum is the same as the integral of the limit as n tends to infinity of the partial sum. Bei Gott what a mouthful! Finally we can exchange the limits! Thank you real analysis.

    ReplyDelete
  6. Wow, actually justifying your interchange of integrals and infinite sums? What kind of engineer are you???
    We'll make a proper mathematician out of you yet papa.

    (Also I didn't know the specification of the DCT so that was really cool)

    ReplyDelete
  7. Didn't you kick away the dct in a video?

    ReplyDelete

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