Better Integration by Parts

April 28, 2023April 30, 2023 / Michael Xue / Leave a comment

From “Integration by Parts Done Right“, we learned

$\int u' \cdot v \;dx = u\cdot v -\int u\cdot v' \; dx.\quad\quad\quad(1)$

It is equally true that

$\int u' \cdot v \;dx = \underline{(u + C)}\cdot v -\int \underline{(u+C)}\cdot v' \; dx\quad\quad\quad(2)$

for any constant $C$ ( see Exercise-1).

Notice (1) is a special case of (2) when $C=0.$

I have observed in many cases when proper choice of $C$ is made, $\int (u+C)\cdot v'\;dx$ can be tremendously simplified.

For example, to evaluate $\int \log(x+k)\; dx,$ we let $C=k$ so

$\int \log(x+k)\;dx = \int x'\cdot \log(x+k)\;dx$

$\overset{(2)}{=} (x+k) \log(x+k)-\int (x+k)\cdot (\log(x+k))'\;dx$

$= (x+k)\log(x+k) -\int (x+k)\cdot \frac{1}{x+k}\;dx$

$= (x+k)\log(x+k) - \int \;dx$

$= (x+k)\log(x+k)-x$

Had we chosen $C=0,$ we would have

$\int \log(x+k)\;dx = \int x'\cdot \log(x+k)\;dx$

$\overset{(2)}{=} x \log(x+k)-\int x\cdot (\log(x+k))'\;dx$

$= x\log(x+k) -\int x\cdot \frac{1}{x+k}\;dx$

$= x\log(x+k) - \int \frac{x+k-k}{x+k}\;dx$

$= x\log(x+k)-\int 1-\frac{k}{x+k}\;dx$

$= x\log(x+k)-x+k\int \frac{1}{x+k}\;dx$

$= x\log(x+k)-x+k\log(x+k)$

$= (x+k)\log(x+k)-x$

Now let’s look at $\int \arctan(\sqrt{x+k})\;dx.$

$\int \arctan(\sqrt{x+k})\;dx = \int x' \arctan(\sqrt{x+k})\;dx$

$\overset{(2)}{=} (x+\underbrace{k+1}_{C})\arctan(\sqrt{x+k})-\int (x+\underbrace{k+1}_{C})\cdot(\arctan(\sqrt{x+k}))'\;dx$

$= (x+k+1)\arctan(\sqrt{x+k})-\int (x+k+1)\cdot \frac{1}{1+(\sqrt{x+k})^2}\cdot\frac{1}{2\sqrt{x+k}}\;dx$

$= (x+k+1)\arctan(\sqrt{x+k})-\int (x+k+1)\cdot \frac{1}{1+x+k}\cdot\frac{1}{2\sqrt{x+k}}\;dx$

$= (x+k+1)\arctan(\sqrt{x+k})-\int \frac{1}{2\sqrt{x+k}}\;dx$

$= (x+k+1)\arctan(\sqrt{x+k})-\sqrt{x+k}$

If we choose $C=k,$

$\int \arctan(\sqrt{x+k})\;dx = \int x' \arctan(\sqrt{x+k})\; dx$

$\overset{(2)}{=} (x+k)\arctan(\sqrt{x+k})-\int (x+k)\frac{1}{1+x+k}\cdot\frac{1}{2\sqrt{x+k}}\;dx$

$= (x+k)\arctan(\sqrt{x+k})-\frac{1}{2}\int\frac{\sqrt{x+k}}{x+k+1}\;dx$

$u = \sqrt{x+k} \implies x=u^2-k, \frac{dx}{du} = 2u$

$= (x+k)\arctan(\sqrt{x+k})-\frac{1}{2}\int\frac{2u^2}{u^2+1}\;du$

$= (x+k)\arctan(\sqrt{x+k})-\frac{1}{2}\int 2(\frac{u^2+1-1}{u^2+1})\;du$

$= (x+k)\arctan(\sqrt{x+k})-\int (1-\frac{1}{u^2+1})\;du$

$= (x+k)\arctan(\sqrt{x+k})-(u-\arctan(u))$

$= (x+k)\arctan(\sqrt{x+k})-(\sqrt{x+k}-\arctan(\sqrt{x+k}))$

$= (x+k+1)\arctan(\sqrt{x+k})-\sqrt{x+k}.$

Fig. 1 Omega‘s choice: $C=k$

It takes longer manually if $C=0$ (see Exercise-3)

To illustrate Just-in-Time determination of $C,$ we evaluate $\int x \arctan(x)\; dx:$

$\int x \arctan(x) \;dx = \int(\frac{x^2}{2})'\arctan(x)\;dx$

$= (\frac{x^2}{2} + C)\arctan(x) - \int (\frac{x^2}{2} + C)(\arctan(x))'\; dx$

$= (\frac{x^2}{2} + C)\arctan(x) - \int (\frac{x^2}{2}+C)\cdot \frac{1}{1+x^2}\; dx$

While the last integral is not difficult we see by choosing $C=\frac{1}{2},$ it is greatly simplified.

$= (\frac{x^2}{2} + \frac{1}{2})\arctan(x) - \frac{1}{2}\int\; dx$

$= (\frac{x^2}{2} + \frac{1}{2}) \arctan(x) - \frac{1}{2}x$

Fig. 2 Omega‘s choice: $C=0$

Exercise-1 Show that $\int u' \cdot v \;dx = \underline{(u + C)}\cdot v -\int \underline{(u+C)}\cdot v' \; dx$ where $C$ is a constant.

Exercise-2 Evaluate $\int \log(\sqrt{x+k})\;dx.$

Exercise-3 Evaluate $\int \frac{x}{(x+k+1)\sqrt{x+k}}\;dx.$

More Extraordinary Sums

April 27, 2023April 28, 2023 / Michael Xue / Leave a comment

The summation derived in my previous post, namely

$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots + \frac{1}{k^2} + \dots,\quad\quad\quad(1)$

is Leonhard Euler’s solution to the problem of finding the precise summation for the reciprocals of squared natural numbers.

We now proceed to find the precise summation for the reciprocals of fourth powered natural numbers:

$1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \dots + \frac{1}{k^4} + \dots$

Recall Euler’s key equation in finding (1):

$1+\frac{x^2}{3!}+\frac{x^4}{5!} + \frac{x^6}{7!} +\frac{x^7}{9!} +\cdots = (1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})(1-\frac{x^2}{16\pi^2})\cdots,\quad\quad\quad(2)$

whose right hand side can be expressed as

$\prod\limits_{i=1}^{n}(1-c_ix^2)\quad\quad\quad(3)$

where $n=1,2,3,..., c_i = \frac{1}{(i\pi)^2}.$ Computer algebra exploration (see Fig. 1) suggests that for finite value $n,$ its coefficient of $x^4$ is

$\sum\limits_{i=1}^{n}\left(c_i\sum\limits_{j=i+1}^{n}c_j\right).$

Fig. 1

Fig. 2

Since

$\sum\limits_{i=1}^{n}\left(c_i\sum\limits_{j=i+1}^{n}c_j\right)=\frac{1}{2}\left((c_1+c_2+c_3+\dots+c_n)^2-(c_1^2+c_2^2+c_3^2+\dots+c_n^2)\right)$

holds for $n=1,2,3,....$ (see (*)), the coefficient of $x^4$ on the right hand side of (2) becomes

$\frac{1}{2}\left((c_1+c_2+c_3+\dots)^2-(c_1^2+c_2^2+ c_3^2 + \dots)\right)$

$\overset{c_i=\frac{1}{(i\pi)^2}}{=}\frac{1}{2}\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi} + \dots)^2-(\frac{1}{\pi^4}+\frac{1}{16\pi^4} + \frac{1}{81\pi^4} + \frac{1}{256\pi^4} + \dots)\right)$

$= \frac{1}{2}\left(\frac{1}{\pi^4}\left(\boxed{1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots}\right)^2-\frac{1}{\pi^4}(1+\frac{1}{16}+\frac{1}{81}+\frac{1}{256}+\dots)\right)$

$\overset{(1)}{=} \frac{1}{2}\left(\frac{1}{\pi^4}\left(\boxed{\frac{\pi^2}{6}}\right)^2-\frac{1}{\pi^4}(1+\frac{1}{16}+\frac{1}{81}+\frac{1}{256}+\dots)\right)$

$= \frac{1}{2}\left(\frac{1}{\pi^4}\frac{\pi^4}{6^2}-\frac{1}{\pi^4}(1+\frac{1}{16}+\frac{1}{81}+\frac{1}{256}+\dots)\right)$

$=\frac{1}{2}\left(\frac{1}{36}-\frac{1}{\pi^4}(1+\frac{1}{16}+\frac{1}{81}+\frac{1}{256}+\dots)\right)$

$= \frac{1}{72}-\frac{1}{2\pi^4}\left(1+\frac{1}{16}++\frac{1}{81} + \frac{1}{256} + \cdots\right).$

Equate the coefficients of $x^4$ on both sides of (2), we have

$\frac{1}{5!} = \frac{1}{72} - \frac{1}{2\pi^4}\left(\underline{1+\frac{1}{16} + \frac{1}{81}+ \frac{1}{256}+ \dots}\right)$

i.e.,

$-\frac{1}{180} = - \frac{1}{2\pi^4}\left(\underline{1+\frac{1}{16} + \frac{1}{81}+ \frac{1}{256}+ \dots}\right)$

Cross multiply $-2\pi^4$ then yields

$1 + \frac{1}{16} + \frac{1}{81} + \frac{1}{256} + \frac{1}{625} + \frac{1}{1296} + \dots + \frac{1}{k^4}+ ...= \frac{\pi^4}{90}.$

Using a Computer Algebra System, we can obtain the precise summation for the reciprocals of even powered natural numbers endlessly.

Fig. 3

A question arises naturally at this point:

For odd integer $i$ , what is $1 +\frac{1}{2^i} +\frac{1}{3^i} + \frac{1}{5^i} + \dots$ ?

We know $1+\frac{1}{2^1}+\frac{1}{3^1} + \frac{1}{4^1} + ...= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ is divergent (see “My shot at Harmonic Series“)

For all other odd values of $i$ , it is convergent (see Exercise-1)

But can we evaluate them precisely?

Some have conjectured that the sum in question is of the form $\left(\frac{p}{q}\right)\cdot\pi^i$ for some fraction $\frac{p}{q}$ as well. But to this day, no one knows if this is true.

The past two hundred years of mathematical research have not advanced our knowledge on the odd powers.

Prove: $\left(\sum\limits_{i=1}^{n}c_i\right)^2-\sum\limits_{i=1}^{n}c_i^2 = 2\sum\limits_{i=1}^{n-1}\left(c_i\sum\limits_{j=i+1}^{n}c_j\right)\quad\quad\quad(*)$

For $n = 2, (*)$ is true:

$\left(\sum\limits_{i=1}^{2}c_i\right)^2-\sum\limits_{i=1}^{2}c_i^2 = (c_1+c_2)^2-(c_1^2+c_2^2) = c_1^2+2c_1c_2+c_2^2-c_1^2-c_2^2 = 2c_1c_2.$

$2\sum\limits_{i=1}^{1}\left(c_i\sum\limits_{j=i+1}^{2}c_j\right) =2c_1\sum\limits_{j=2}^{2}c_j = 2c_1c_2$

Assume for $n=k, (*)$ is true:

$\left(\sum\limits_{i=1}^{k}c_i\right)^2-\sum\limits_{i=1}^{k}c_i^2 = 2\sum\limits_{i=1}^{k-1}\left(c_i\sum\limits_{j=i+1}^{k}c_j\right).\quad\quad\quad(A-1)$

Let’s show (*) is true when $n=k+1$ :

$\left(\sum\limits_{i=1}^{k+1}c_i\right)^2-\sum\limits_{i=1}^{k+1}c_i^2$

$= \left(\sum\limits_{i=1}^{k}c_i +c_{k+1}\right)^2-\sum\limits_{i=1}^{k+1}c_i^2$

$= \left(\sum\limits_{i=1}^{k}c_i\right)^2 + 2 (\sum\limits_{i=1}^{k}c_i)c_{k+1} + c_{k+1}^2-\sum\limits_{i=1}^{k}c_i^2 - c_{k+1}^2$

$= \left(\sum\limits_{i=1}^{k}c_i\right)^2 + 2 \sum\limits_{i=1}^{k}(c_i c_{k+1}) -\sum\limits_{i=1}^{k}c_i^2$

$= \underline{\left(\sum\limits_{i=1}^{k}c_i\right)^2 -\sum\limits_{i=1}^{k}c_i^2} + 2 \sum\limits_{i=1}^{k}(c_i c_{k+1})$

$\overset{(A-1)}{=} 2\sum\limits_{i=1}^{k-1}\left(c_i\sum\limits_{j=i+1}^{k}c_j\right) + 2 \sum\limits_{i=1}^{k}(c_i c_{k+1})$

$= 2\left(\sum\limits_{i=1}^{k-1}\left(c_i\sum\limits_{j=i+1}^{k}c_j\right) + \sum\limits_{i=1}^{k}(c_i c_{k+1})\right)$

$= 2\biggl(c_1(c_2+c_3+\dots+c_k) +c_1c_{k+1} + c_2(c_3+c_4 + \dots + c_k) + c_2c_{k+1}+\dots+ c_{k-2}(c_{k-1} + c_k) + c_{k-2}c_{k+1}+ c_{k-1}c_k +c_{k-1}c_{k+1}+ c_kc_{k+1}\biggr)$

$= 2\biggl(c_1(c_2+c_3 + \dots + c_{k+1})+ c_2(c_3+c_4 + \dots + c_{k+1})+\dots+ c_{k-2}(c_{k-1} + c_k + c_{k+1}) + c_{k-1}(c_k+c_{k+1}) + c_k c_{k+1}\biggr)$

$= 2\sum\limits_{i=1}^{k}\left(c_i\sum\limits_{j=i+1}^{k+1}c_j\right).$

$= 2\sum\limits_{i=1}^{(k+1)-1}\left(c_i\sum\limits_{j=i+1}^{k+1}c_j\right).$

Exercise-1 Show that for $i \ge 2, \sum\limits_{k=1}^{\infty}\frac{1}{k^i}$ is convergent.

Exercise-2 Derive $\sum\limits_{k=1}^{\infty}\frac{1}{k^i}$ for $i = 6, 8, 10, ..., 30.$

Deriving the Extraordinary Euler Sum

April 17, 2023June 20, 2023 / Michael Xue / Leave a comment

By definition,

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} + \dots$

Dividing by $x \ne 0$ yields

$\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \frac{x^8}{9!}+\dots\quad\quad\quad(1)$

Assuming the properties of finite polynomials hold true for infinite series,

$\sin(x) = 0$ when $x=\pm \pi, \pm 2\pi, \pm 3\pi, ...$

means

$1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \frac{x^8}{9!}+\dots$

$= (1-\frac{x}{\pi})(1-\frac{x}{-\pi})(1-\frac{x}{2\pi})(1-\frac{x}{-2\pi})(1-\frac{x}{3\pi})(1-\frac{x}{-3\pi})\dots$

$= (1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\pi})(1+\frac{x}{2\pi})(1-\frac{x}{3\pi})(1+\frac{x}{3\pi})\dots$

$=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})(1-\frac{x^2}{16\pi^2})\dots$

i.e.,

$1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \frac{x^8}{9!}+\dots=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})(1-\frac{x^2}{16})\dots\quad\quad\quad(2)$