Integration using Beta and Gamma Functions

[alyafey22]

The following sum diverges

$$\sum_{k\geq 1}\frac{H_k}{k}$$

Actually the Euler number $\gamma$ can be defined as

$$\gamma = \lim_{k\to \infty}H_k-\log(k)$$

So Harmonic number differ by a constant term as they approach infinity.

Hence we can write

$$\sum_{k\geq 1}\frac{H_k}{k} \sim \sum_{k\geq 1}\frac{\log(k)}{k} $$

The right integral diverges by the integral test.

You don't have to understand all of that but you have to know that diverges.

So the equation

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

only converges for $q \geq 2$ and by definition for $q=2$ we have

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)-\frac{1}{2}\sum_{k=1}^{0}\zeta(k+1)\zeta(2-k)$$

The sum is by definition equal to 0 so

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)$$

For references you can search for Euler sums , you will find lots of papers.

 

[Amad27]

The following sum diverges

$$\sum_{k\geq 1}\frac{H_k}{k}$$

Actually the Euler number $\gamma$ can be defined as

$$\gamma = \lim_{k\to \infty}H_k-\log(k)$$

So Harmonic number differ by a constant term as they approach infinity.

Hence we can write

$$\sum_{k\geq 1}\frac{H_k}{k} \sim \sum_{k\geq 1}\frac{\log(k)}{k} $$

The right integral diverges by the integral test.

You don't have to understand all of that but you have to know that diverges.

So the equation

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

only converges for $q \geq 2$ and by definition for $q=2$ we have

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)-\frac{1}{2}\sum_{k=1}^{0}\zeta(k+1)\zeta(2-k)$$

The sum is by definition equal to 0 so

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^2}=2\zeta(3)$$

For references you can search for Euler sums , you will find lots of papers.

But the issue is we don't have an $n^2$ term when evaluating the $k+1$ sum. Please let me know if there is a work around for this.
Meanwhilw, I just had an idea.

We can consider the integral definition of the Harmonix number.

$$H_k = \int_{0}^{1} (1-x)^{k-1} dx$$

And convert the whole sum.

I used integral on the original geometric series formula and I got an intense formula involving several poly logarithms. Perhaps this is a way? Will you be interested on seeing that attempt?

 

[alyafey22]

But the issue is we don't have an $n^2$ term when evaluating the $k+1$ sum. Please let me know if there is a work around for this.

I don't understand what you mean .

Meanwhilw, I just had an idea.

We can consider the integral definition of the Harmonix number.

$$H_k = \int_{0}^{1} (1-x)^{k-1} dx$$

And convert the whole sum.

I used integral on the original geometric series formula and I got an intense formula involving several poly logarithms. Perhaps this is a way? Will you be interested on seeing that attempt?

Of course , let me see if you got any other attempts. It might be better than my approach , why not ?

 

[Amad27]

I don't understand what you mean .
Of course , let me see if you got any other attempts. It might be better than my approach , why not ?

So we consider just

$$I = \sum_{k=1}^{\infty}\frac{\int_{0}^{1} (1-x)^{k-1} \,dx}{(k+1)(k+2)^3}$$

$$I = \int_{0}^{1} \sum_{k=1}^{\infty} \frac{(1-x)^{k-1}}{(k+1)(k+2)^3} \,dx$$

Now it is about to get extremely complicated. Principally, we know

$$\sum_{k=1}^{\infty} x^{k-1} = \frac{1}{1-x}$$

$$\sum_{k=1}^{\infty} (1-x)^{k-1} = \frac{1}{x}$$

$$\sum_{k=1}^{\infty} (1-x)^{k} = \frac{(1-x)}{x}$$

$$\sum_{k=1}^{\infty} \frac{(1-x)^{k+1}}{k+1} = x - \log(x)$$

$$\sum_{k=1}^{\infty} \frac{(1-x)^{k+2}}{(k+1)(k+2)} = \frac{x^2}{2} + x - x\log(x)$$

$$\sum_{k=1}^{\infty} \frac{(1-x)^{k+1}}{(k+1)(k+2)} = \frac{\frac{x^2}{2} + x - x\log(x)}{(1-x)}$$

I need help here:

I can't integrate because I checked on W|A and there is a dilogarithmic term involved.

How do you express polylogs as indefnite integrals? So that there is no upper or lower bound?

From what I know:

$$Li_2(x) = (-)\cdot\int \frac{\log(x)}{(1-x)}$$
$$Li_2(1-x) = (-)\cdot \int \frac{\log(1-x)}{x}$$

Perhaps we could use integration by parts to get this factor in there?

There is always a question I've had:

How can you define a function $f(a)$ such that $a$ is actually inside the integral? So:

$$f(a) = \int ax^2 dx$$

Usually, textbooks define integral functions as the variable being the upper limit of a definite integral.

So what property/rule of integral, or definition states you can have a function of a variable inside the integral itself? Thanks!

 

[alyafey22]

$$H_k = \int_{0}^{1} (1-x)^{k-1} dx$$

Sorry , I missed the definition ... Why is that true ?

 

[Amad27]

Sorry , I missed the definition ... Why is that true ?

Oh oh...

That's why it wasn't working.

I used the wrong definition. It is

$$H_k = \int_{0}^{1} \frac{1 - x^k}{1-x} \,dx$$

The $(1-x)$ isn't in brackets...

Please give me some time to think over this. Meanwhile... can you answer the other question I asked?

Question:
How can you define a function $f(a)$ such that $a$ is actually inside the integral? So:

$$f(a) = \int ax^2 dx$$

Usually, textbooks define integral functions as the variable being the upper limit of a definite integral.

So what property/rule of integral, or definition states you can have a function of a variable inside the integral itself? Thanks!

 

[alyafey22]

Question:
How can you define a function $f(a)$ such that $a$ is actually inside the integral? So:

$$f(a) = \int ax^2 dx$$

Usually, textbooks define integral functions as the variable being the upper limit of a definite integral.

So what property/rule of integral, or definition states you can have a function of a variable inside the integral itself? Thanks!

You can do the following

\(\displaystyle f(a) = \int^{t_1}_{t_0} g(t,a)\,dt\)

where $t_0,t_1$ are constants.

Indeed , it is just a symbol to define the integral. For example we have

\(\displaystyle \Gamma(a) = \int^\infty_0 t^{a-1} \, e^{-t} \, dt\)

where in this case $f(a) = \Gamma(a) $ and $g(t,a) = t^{a-1} \, e^{-t} $.

Since $a$ is independent of $t$ we can define something like this

$$g(a) = a \int^1_0 f(t) \, dt $$

I don't recall a name for this property but is mainly used to differentiate w.r.t to the variable inserted.

 

[Amad27]

You can do the following

\(\displaystyle f(a) = \int^{t_1}_{t_0} g(t,a)\,dt\)

where $t_0,t_1$ are constants.

Indeed , it is just a symbol to define the integral. For example we have

\(\displaystyle \Gamma(a) = \int^\infty_0 t^{a-1} \, e^{-t} \, dt\)

where in this case $f(a) = \Gamma(a) $ and $g(t,a) = t^{a-1} \, e^{-t} $.

Since $a$ is independent of $t$ we can define something like this

$$g(a) = a \int^1_0 f(t) \, dt $$

I don't recall a name for this property but is mainly used to differentiate w.r.t to the variable inserted.

I suppose it is a property of integral functions. Anywho.

I worked on the problem, and found out how to do it... until a point.

We understand

$$I = \int_{0}^{1} \left(\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} + \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^2}\right) \, dx$$

Assuming both of these sums are convergent. I used the formula,

$$\sum_{k=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$ to get a formula for a general $x^{k+2}$.

$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -Li_3(x) - Li_2(x) - \frac{x^2}{8} + 3x - x\log(1-x) + \log(1-x)$$

If you want to see the workings, I can try to scan it, because it took a few pages... (I used W|A for some integrals, instead of doing integration by parts).

Letting $x=1$, you get

$$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} = -Li_3(1) - Li_2(1) - \frac{1}{8} + 3 - \log(0) + \log(0)$$

Here is the crises, $\log(0)$ is undefined, is it okay to subtract this: $\text{undefined} - \text{undefined}$ ?? Let me know of this property/theorem.

$$= -\zeta(3) - \zeta(2) + \frac{23}{8}$$

I am working on the other part, but was the logarithmic subtraction correct?

Thanks

 

[alyafey22]

$$I = \int_{0}^{1} \left(\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} + \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^2}\right) \, dx$$

I don't seem to understand how you got that!

You used

$$H_k = \int^1_0 \frac{1-x^k}{1-x}\,dx $$

right ?

 

[Amad27]

I don't seem to understand how you got that!

You used

$$H_k = \int^1_0 \frac{1-x^k}{1-x}\,dx $$

right ?

FIRST A QUESTION:
I asked this before, but can we do $\log(0) - \log(0) = 0$? Otherwise this idea doesn't work.

Now, the workings:

I'll try to do the best I can.

$$H_k = \int^1_0 \frac{1-x^k}{1-x}\,dx $$
We had

$$(2)\cdot\int_{0}^{1} \frac{1}{1-x}\left(\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} - \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^3}\right) \,dx$$

Through Fubini's theorem, you interchange the sum and integral, then $1/(1-x)$ is just a constant for the sum, so you can "extract," that out. Then split the sum, ASSUMING both are convergent. Since I am too lazy to do a test. Actually I don't know how to test a series for convergence anyway...

Then the sum of

$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -[Li_3(x) + Li_2(x)] - x^2/8 + 3x - x\log(1-x) + \log(1-x)$$

Letting $x=1$, the sum is: $S_1 = -\zeta(3) - \zeta(2) + 23/8$

Dividing the sum of $x^{k+2}$ by $x^2$ we get.

$$\sum_{k=1}^{\infty} \frac{x^{k}}{(k+1)(k+2)^3} = -[Li_3(x) + Li_2(x)]/x^2 - [x^2/8 + 3x - x\log(1-x) + \log(1-x)]/x^2$$

We had

$$(2)\cdot\int_{0}^{1} \frac{1}{1-x}\left(\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+2)^3} - \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^3}\right) \,dx$$

$$= (2)\cdot\int_{0}^{1} \frac{1}{1-x}\left(-\zeta(3) - \zeta(2) + 23/8 - \left(-[Li_3(x) + Li_2(x)]/x^2 - [x^2/8 + 3x - x\log(1-x) + \log(1-x)]/x^2\right)\right) \,dx$$

$$= (2)\cdot\int_{0}^{1} \frac{-\zeta(3) - \zeta(2) + 23/8}{1-x}\left( - \left(\frac{-Li_3(x) - Li_2(x) - \frac{x^2}{8} - 3x + x\log(1-x) - \log(1-x)}{(1-x)x^2}\right)\right) \,dx$$

$$ = (2)\cdot\int_{0}^{1} \frac{-x^2\zeta(3) - x^2\zeta(2) + \frac{23x^2}{8} + Li_3(x) + Li_2(x) + \frac{x^2}{8} + 3x - x\log(1-x) + \log(1-x)}{(1-x)x^2} \,dx$$

I think we should split this, but this integral product should give the final answer.

But how come $\displaystyle \int_{0}^{1} \frac{Li_3(x) + Li_2(x)}{(1-x)x^2} \,dx$ does NOT converge?

 

[alyafey22]

$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -[Li_3(x) + Li_2(x)] - x^2/8 + 3x - x\log(1-x) + \log(1-x)$$

Let us go step by step first. How did you prove the above formula ?

 

[Amad27]

Let us go step by step first. How did you prove the above formula ?

I will post it the second I get home. But for now,

How can I check if an integral converges or not?

I tried the integral

$$\int_{0}^{1} \frac{-Li_3(x) - Li_2(x)}{(1-x)x^2} \,dx$$

But it does not converge. How can I know if the integral, will converge or not BEFOREhand.QUESTION 2: Which types of integrals involving Polylogs converge?

Thanks =)

 

[alyafey22]

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1-x}\,dx$$

Does not converge. This can be see using

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1-x}\,dx = \int^1_0 \frac{\mathrm{Li}_2(1-x)}{x}\,dx$$

Then

$$\mathrm{Li}_2(1-x) = \zeta(2) -\mathrm{Li}_2(x)-\log(x)\log(1-x)$$

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1-x}\,dx = \int^1_0 \frac{\zeta(2) -\mathrm{Li}_2(x)-\log(x)\log(1-x)}{x}\,dx $$

First integral

$$\int^1_0 \frac{\log(x)\log(1-x)}{x}\,dx = \int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx $$

The following integral

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx$$

converges since , function to integrate is continuous so the integral converges.

It remains

$$\int^1_0 \frac{\zeta(2)}{x}\,dx$$

is clearly divergent.

There are many ways to apply , but generally you have to do some change of variable or integration by part to see why the integral diverges. Sometimes , you have to do some computations.

In other cases , you can bound the integral. For example

$$\int^\infty_1 \frac{e^{-x}}{x^2}\,dx$$

How to prove that the improper integral converges ?

$$|\int^\infty_1 \frac{e^{-x}}{x^2}\,dx| \leq \int^\infty_1 | \frac{e^{-x}}{x^2}| \,dx \leq \int^\infty_1 \frac{1}{x^2} < \infty$$

Since the area of the integral is less than a finite area , the integral converges.

 

[Amad27]

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1-x}\,dx$$

Does not converge. This can be see using

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1-x}\,dx = \int^1_0 \frac{\mathrm{Li}_2(1-x)}{x}\,dx$$

Then

$$\mathrm{Li}_2(1-x) = \zeta(2) -\mathrm{Li}_2(x)-\log(x)\log(1-x)$$

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1-x}\,dx = \int^1_0 \frac{\zeta(2) -\mathrm{Li}_2(x)-\log(x)\log(1-x)}{x}\,dx $$

First integral

$$\int^1_0 \frac{\log(x)\log(1-x)}{x}\,dx = \int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx $$

The following integral

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx$$

converges since , function to integrate is continuous so the integral converges.

It remains

$$\int^1_0 \frac{\zeta(2)}{x}\,dx$$

is clearly divergent.

There are many ways to apply , but generally you have to do some change of variable or integration by part to see why the integral diverges. Sometimes , you have to do some computations.

In other cases , you can bound the integral. For example

$$\int^\infty_1 \frac{e^{-x}}{x^2}\,dx$$

How to prove that the improper integral converges ?

$$|\int^\infty_1 \frac{e^{-x}}{x^2}\,dx| \leq \int^\infty_1 | \frac{e^{-x}}{x^2}| \,dx \leq \int^\infty_1 \frac{1}{x^2} < \infty$$

Since the area of the integral is less than a finite area , the integral converges.

Here's the proof

$$\sum_{k=1}^{\infty} x^{k-1} = \frac{1}{1-x}$$
$$\sum_{k=1}^{\infty} x^{k} = \frac{x}{1-x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{k+1} = -x - \log(1-x)$$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)} = \frac{-x^2}{2} + x - x\log(1-x) + \log(1-x)$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{(k+1)(k+2)} = \frac{-x}{2} + 1 - \log(1-x) + \frac{\log(1-x)}{x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^2} = \log(1-x) - x\log(1-x) + 2x - \frac{x^2}{4} - Li_2(x) $$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{(k+1)(k+2)^2} = \frac{\log(1-x)}{x} - \log(1-x) + 2 - \frac{x}{4} - \frac{Li_2(x)}{x} $$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -Li_2(x)+ x - x\log(1-x) + \log(1-x) + 2x - \frac{x^2}{8} - Li_3(x) \blacksquare$$

 

[alyafey22]

Here's the proof

$$\sum_{k=1}^{\infty} x^{k-1} = \frac{1}{1-x}$$
$$\sum_{k=1}^{\infty} x^{k} = \frac{x}{1-x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{k+1} = -x - \log(1-x)$$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)} = \frac{-x^2}{2} + x - x\log(1-x) + \log(1-x)$$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{(k+1)(k+2)} = \frac{-x}{2} + 1 - \log(1-x) + \frac{\log(1-x)}{x}$$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^2} = \log(1-x) - x\log(1-x) + 2x - \frac{x^2}{4} - Li_2(x) $$
$$\sum_{k=1}^{\infty} \frac{x^{k+1}}{(k+1)(k+2)^2} = \frac{\log(1-x)}{x} - \log(1-x) + 2 - \frac{x}{4} - \frac{Li_2(x)}{x} $$
$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -Li_2(x)+ x - x\log(1-x) + \log(1-x) + 2x - \frac{x^2}{8} - Li_3(x) \blacksquare$$

Seems correct. Let us verify it by taking the limit as $x \to 1$.

What do you get ?

Does it agree with the value returned by wolfram.

 

[Amad27]

Seems correct. Let us verify it by taking the limit as $x \to 1$.

What do you get ?

Does it agree with the value returned by wolfram.

The issue is with taking the limit.

$\displaystyle lim_{x\to1} \log(1-x)$ does not exist.

 

[alyafey22]

$$\sum_{k=1}^{\infty} \frac{x^{k+2}}{(k+1)(k+2)^3} = -Li_2(x)+ x - x\log(1-x) + \log(1-x) + 2x - \frac{x^2}{8} - Li_3(x) $$

Take limits of both sides. Don't distribute the limit until you are sure the limit exists.

 

[Amad27]

Take limits of both sides. Don't distribute the limit until you are sure the limit exists.

$$\lim_{{x}\to{1}} -Li_2(x)+ x - x\log(1-x) + \log(1-x) + 2x - \frac{x^2}{8} - Li_3(x)$$
$$= \lim_{{x}\to{1}} -Li_2(x) +(\log(1-x))(1 - x) + 3x - \frac{x^2}{8} - Li_3(x)$$

The more interesting case is $\log(1-x)\cdot(1-x)$

$$y = \lim_{{x}\to{1}} (\log(1-x)\cdot(1-x))$$
Let $u = 1-x$ as $x\to1$ we have $u \to 0$

$$y = \lim_{{u}\to{0}} \log(u)\cdot(u) = \lim_{{u}\to{0}} \log(u\cdot e^u)$$

$$e^y = \lim_{{u}\to{0}} 0$$

But then we are back to where we started with $\log(0)$

 

[alyafey22]

$$\lim_{u \to 0} u \times \log(u) = \lim_{u \to 0} \frac{\log(u)}{\frac{1}{u}} $$

By L'Hospital rule the limit goes to 0.

 

[Amad27]

$$\lim_{u \to 0} u \times \log(u) = \lim_{u \to 0} \frac{\log(u)}{\frac{1}{u}} $$

By L'Hospital rule the limit goes to 0.

That is very nice, but the denominator is not indeterminate is it? $1/0$ is not indeterminate is it?

$$= \lim_{{x}\to{1}} -Li_2(x) +(\log(1-x))(1 - x) + 3x - \frac{x^2}{8} - Li_3(x)$$

$$= \lim_{{x}\to{1}} -Li_2(x) + \frac{23}{8} - Li_3(x)$$

$$= -\zeta(2) + \frac{23}{8} - \zeta(3)$$

Then

$$ \sum_{k=1}^{\infty} \frac{x^k}{(k+1)(k+2)^3} = -\frac{Li_2(x)}{x^2}+ \frac{1}{x} - \frac{\log(1-x)}{x} + \frac{\log(1-x)}{x^2} + \frac{2}{x} - \frac{1}{8} - \frac{Li_3(x)}{x^2}$$

Then we need to integrate:

$$\int_{0}^{1} -\frac{Li_2(x)}{(1-x)x^2}+ \frac{1}{(1-x)x} - \frac{\log(1-x)}{x(1-x)} + \frac{\log(1-x)}{(1-x)x^2} + \frac{2}{(1-x)x} - \frac{1}{8(1-x)} - \frac{Li_3(x)}{(1-x)x^2} \,dx$$

This is the most difficult part of all. Especially because if this is split up, a lot of it does not converge.

out of curiosity: could we use complex analysis to solve the sum we had? Also, can you also help me through some complex analysis, and contour integration? I am quite interested, but I don't have means to learn it except MHB. Help will be appreciated! =)

 

[alyafey22]

That is very nice, but the denominator is not indeterminate is it? $1/0$ is not indeterminate is it?

The limit is of the form $\frac{\infty}{\infty}$. which can be solved by L'Hospital rule.

Then we need to integrate:

$$\int_{0}^{1} -\frac{Li_2(x)}{(1-x)x^2}+ \frac{1}{(1-x)x} - \frac{\log(1-x)}{x(1-x)} + \frac{\log(1-x)}{(1-x)x^2} + \frac{2}{(1-x)x} - \frac{1}{8(1-x)} - \frac{Li_3(x)}{(1-x)x^2} \,dx$$

This is the most difficult part of all. Especially because if this is split up, a lot of it does not converge.

I guess this will be complicated , at least you tried :) . Why don't we stick to the solution using Harmonic numbers ?

out of curiosity: could we use complex analysis to solve the sum we had? Also, can you also help me through some complex analysis, and contour integration? I am quite interested, but I don't have means to learn it except MHB. Help will be appreciated! =)

Of course.I 'll answer this question in the other thread.

 

[Amad27]

The limit is of the form $\frac{\infty}{\infty}$. which can be solved by L'Hospital rule.
I guess this will be complicated , at least you tried :) . Why don't we stick to the solution using Harmonic numbers ?
Of course.I 'll answer this question in the other thread.

But how do we use that formula?

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

Lets consider $n = \displaystyle \sqrt{k+1}$

Then the lower limit will be

$k = 0$

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

Nevermind, I just realized the $H_n$ won't fit properly... Wow, this is difficult.

 

[alyafey22]

Ok , I will give you quick exercise

Find

\(\displaystyle \sum_{k\geq 0}\frac{H_k}{(k+1)^2}\)

 

[Amad27]

Ok , I will give you quick exercise

Find

\(\displaystyle \sum_{k\geq 0}\frac{H_k}{(k+1)^2}\)

Ok.

Let $n = j+1$

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

$$\sum_{j=0}^\infty \frac{H_{j+1}}{(j+1)^2}= \left(1+\frac{2}{2} \right)\zeta(3)-\frac{1}{2}\sum_{k=1}^{0}\zeta(k+1)\zeta(3-k)$$

$$\sum_{j=0}^\infty \frac{H_{j+1}}{(j+1)^2} =
\sum_{j=0}^\infty \frac{H_{j} + \frac{1}{j+1}}{(j+1)^2} = \sum_{j=0}^\infty \frac{H_{j}}{(j+1)^2}+ \sum_{j=0}^{\infty} + \frac{1}{(j+1)^3}$$

Originally, the formula states:

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

$$\sum_{n=1}^\infty \frac{H_n}{n^q} = \sum_{n=0}^\infty \frac{H_n}{n^q} - 1$$

Then

$$\sum_{n=1}^\infty \frac{H_j}{(j+1)^2} = \sum_{n=0}^\infty \frac{H_j}{(j+1)^2} - 1 = (2)\zeta(3) - \frac{1}{2}(indeterminate)$$

I checked W|A, the sum involving the $q-2$ is indeterminate, which is bad news.

 

[alyafey22]

Ok.

Let $n = j+1$

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

$$\sum_{j=0}^\infty \frac{H_{j+1}}{(j+1)^2}= \left(1+\frac{2}{2} \right)\zeta(3)-\frac{1}{2}\sum_{k=1}^{0}\zeta(k+1)\zeta(3-k)$$

$$\sum_{j=0}^\infty \frac{H_{j+1}}{(j+1)^2} =
\sum_{j=0}^\infty \frac{H_{j} + \frac{1}{j+1}}{(j+1)^2} = \sum_{j=0}^\infty \frac{H_{j}}{(j+1)^2}+ \sum_{j=0}^{\infty} + \frac{1}{(j+1)^3}$$

Originally, the formula states:

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

$$\sum_{n=1}^\infty \frac{H_n}{n^q} = \sum_{n=0}^\infty \frac{H_n}{n^q} - 1$$

Then

$$\sum_{n=1}^\infty \frac{H_j}{(j+1)^2} = \sum_{n=0}^\infty \frac{H_j}{(j+1)^2} - 1 = (2)\zeta(3) - \frac{1}{2}(indeterminate)$$

I checked W|A, the sum involving the $q-2$ is indeterminate, which is bad news.

By definition if the upper index of the sum is 0 then the sum is 0.

 

[Amad27]

By definition if the upper index of the sum is 0 then the sum is 0.

Then why does W|A return it as indeterminate?

 

[alyafey22]

Then why does W|A return it as indeterminate?

W|A can't solve it because the lower index is greater than the upper index.

So we have to add the following constrains $q>1$ and when $q=2$ the series is equal to 0. It is just a constrain that we defined on that particular case.

Any way, Let us assume that we can't find the summation when $q=2$ so we have to find

$$\sum_{k\geq 1} \frac{H_k}{k^2}$$

Can you think of a way to evaluate it without using that formula ?

 

[Amad27]

W|A can't solve it because the lower index is greater than the upper index.

So we have to add the following constrains $q>1$ and when $q=2$ the series is equal to 0. It is just a constrain that we defined on that particular case.

Any way, Let us assume that we can't find the summation when $q=2$ so we have to find

$$\sum_{k\geq 1} \frac{H_k}{k^2}$$

Can you think of a way to evaluate it without using that formula ?

Not really. I suppose the formula is the best way, and I suppose by definition the sum is $0$.

$$\sum_{n=1}^\infty \frac{H_j}{(j+1)^2} = \sum_{n=0}^\infty \frac{H_j}{(j+1)^2} - 1 = (2)\zeta(3) - 0 - 1 = 2\zeta(3) - 1$$

 

[alyafey22]

Hint :

use that

$$\int^1_0 \frac{1-x^n}{1-x}\,dx = H_n$$

 

[Amad27]

Hint :

use that

$$\int^1_0 \frac{1-x^n}{1-x}\,dx = H_n$$

That is what I did a while ago (with the other sum) and it became a mess as you saw.

$$\sum_{k=1}^{\infty} \frac{H_k}{k^2}$$

$$H_k = \int_{0}^{1} \frac{1-x^k}{1-x} \,dx$$

$$S = \int_{0}^{1} \frac{1}{1-x} \sum_{k=1}^{\infty} \frac{1-x^k}{k^2} \,dx$$

Assuming the parts of the sum are convergent we get:

$$S = \int_{0}^{1} \frac{1}{1-x} \left(\zeta(2) - \sum_{k=1}^{\infty} \frac{x^k}{k^2}\right) \,dx$$

$$S = \int_{0}^{1} \frac{1}{1-x} \left(\zeta(2) - Li_2(x)\right) \,dx$$

$$S = \lim_{{x}\to{1}} -\zeta(2)\log(1-x) - \int_{0}^{1} \frac{Li_2(x)}{1-x} \,dx$$

The Li_2(x)/(1-x) is the hard part. Integration by parts or some other strategy?

 

[alyafey22]

The limit doesn't exist so you cannot separate the two integrals because they are divergent as I said previously.

$$Li_2(x)+Li_2(1-x) = \zeta(2)-\log(x)\log(1-x)$$

$$\zeta(2)-Li_2(x) = Li_2(1-x)+\log(x)\log(1-x)$$

 

[Amad27]

The limit doesn't exist so you cannot separate the two integrals because they are divergent as I said previously.

$$Li_2(x)+Li_2(1-x) = \zeta(2)-\log(x)\log(1-x)$$

$$\zeta(2)-Li_2(x) = Li_2(1-x)+\log(x)\log(1-x)$$

$$S = \int_{0}^{1} \frac{Li_2(1-x) + \log(x)\log(1-x)}{1-x} \,dx$$

Assuming this integral is convergent, I suppose can split this.

$$S = -Li_3(1-x)]_{0}^{1} - \text{problem}$$

Okay, this is depressing.

And very hard.

 

[alyafey22]

Try integration by parts , don't quite so easily.

 

[Amad27]

Try integration by parts , don't quite so easily.

NEW IDEA!

$\log(1-x)\log(x) = -Li_2(1-x) - Li_2(x) + \zeta(2)$

$$S = \int_{0}^{1} \frac{-Li_2(x) + \zeta(2)}{1-x} \,dx$$

First the antiderivative:

$$S = (\zeta(2) - Li_2(x)) + \int \frac{\log^2(1-x)}{(1-x)} dx$$

Let $\log(1-x) = u \implies du = \frac{1}{1-x}$

$$S_{antiderivative} = (\zeta(2) - Li_2(x)) + \frac{\log^3(1-x)}{3}$$

$$S = \lim_{{x}\to{1}} -Li_2(x) + \frac{\log^3(1-x)}{3}$$

Wait, does this limit exist?

 

[alyafey22]

NEW IDEA!

$\log(1-x)\log(x) = -Li_2(1-x) - Li_2(x) + \zeta(2)$

$$S = \int_{0}^{1} \frac{-Li_2(x) + \zeta(2)}{1-x} \,dx$$

First the antiderivative:

$$S = (\zeta(2) - Li_2(x)) + \int \frac{\log^2(1-x)}{(1-x)} dx$$

I don't understand how you got that. Moreover , S shouldn't be used twice since the first is definite and the second is indefinite integral.