Integration using Beta and Gamma Functions

[Amad27]

The iteration variable here is $k$ not $n$.
Use the following property

$$\int^x_0 t^{n-1}\,dt = \frac{x^n}{n}$$

From your hints, I believe I am to do the following:

$$\int n{z}^{(n-1)} \,dz = z^n$$

But in any case, $z$ is not the iteration variable, so we can't integrate with respect to $z$ now can we?

 

[alyafey22]

From your hints, I believe I am to do the following:

$$\int n{z}^{(n-1)} \,dz = z^n$$

But in any case, $z$ is not the iteration variable, so we can't integrate with respect to $z$ now can we?

I mean use the series

\(\displaystyle \sum_{n \geq 1}z^{n-1} = \frac{1}{1-z}\)

And integrate with respect to $z$.

 

[Amad27]

I mean use the series

\(\displaystyle \sum_{n \geq 1}z^{n-1} = \frac{1}{1-z}\)

And integrate with respect to $z$.

mmmmm...

Isnt $n$ the iteration variable? $z$ could merely be considered a constant (like in a partial derivative).

Then how can you integrate with respect to $z$ and not $n$?

How do you know, which is an iteration variable?

Thanks!

 

[alyafey22]

The iteration variable is the one that changes in the summation. It might be easier for you if I expanded the summation as

\(\displaystyle \sum_{n\geq 1} z^{n-1} = 1+z+z^2+z^3 +\cdots \)

As we see the iteration variable is $n$ and is the power of $z$. Can you see how we can integrate with respect to $z$. Integrating with respect to $n$ doesn't make sense because it is changing as we iterate.

 

[Amad27]

The iteration variable is the one that changes in the summation. It might be easier for you if I expanded the summation as

\(\displaystyle \sum_{n\geq 1} z^{n-1} = 1+z+z^2+z^3 +\cdots \)

As we see the iteration variable is $n$ and is the power of $z$. Can you see how we can integrate with respect to $z$. Integrating with respect to $n$ doesn't make sense because it is changing as we iterate.

Okay, I'll ask and "do" at the same exact time.

(Q1) So, the iteration variable is sort of like a constant, since in the expansion it is actually a number, which is why we don't integrate with respect to it?

(A1) I'll do the integration as a well;

$$\int z^{n-1} \,dz = \frac{z^n}{n}$$

So we have the $n$ in the denominator,

We need now $\frac{1}{(2n-1)}$ left to work with. In

$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)}$$

 

[alyafey22]

Okay, I'll ask and "do" at the same exact time.

(Q1) So, the iteration variable is sort of like a constant, since in the expansion it is actually a number, which is why we don't integrate with respect to it?

For the time being thing of the iteration variable as an index that helps us find the terms of the summation or sequence. Most of the time it is not of that interest for us. Integrating or differentiating with respect to it doesn't make sense because it is discrete . It only represents a subset of the integers. What we are interested in manipulating is the variable that the iteration is performed on. In that sense we have

$$ \sum_{n\geq 1}\frac{z^n}{n} = - \log(1-z)$$

Now we need to get $(2n-1)$ in the denominator. Can you think of a way to do this ? Always think about integration.

 

[Amad27]

For the time being thing of the iteration variable as an index that helps us find the terms of the summation or sequence. Most of the time it is not of that interest for us. Integrating or differentiating with respect to it doesn't make sense because it is discrete . It only represents a subset of the integers. What we are interested in manipulating is the variable that the iteration is performed on. In that sense we have

$$ \sum_{n\geq 1}\frac{z^n}{n} = - \log(1-z)$$

Now we need to get $(2n-1)$ in the denominator. Can you think of a way to do this ? Always think about integration.

Hello, what does discrete mean? Secondly, We can't integrate anything to get $(2n - 1)$ in the denominator because it is supposedly a constant isn't it?

Thanks! A lot!

 

[alyafey22]

Hello, what does discrete mean? Secondly,

Not continuous.

We can't integrate anything to get $(2n - 1)$ in the denominator because it is supposedly a constant isn't it?

Thanks! A lot!

What do you mean by constant ? We should integrate with respect to $z$ , right ?

 

[Amad27]

Not continuous.
What do you mean by constant ? We should integrate with respect to $z$ , right ?

First then, how is an iteration variable non continuous? Because it change like $n = 1,2,3,4,5...$?? Perhaps?

Then secondly, I didnt see that $log$ sorry!

We had

$-\ln(1-z) = u'$
$u = -\frac{1}{1-z}$

We have a fraction successfully, all we need now is the $n$.

We can multiply and divide by $n$

$$u = -\frac{2n}{2n - n} = -\frac{2}{1}$$

Never mind, this won't work.

 

[alyafey22]

First then, how is an iteration variable non continuous? Because it change like $n = 1,2,3,4,5...$?? Perhaps?

Yes , as I said it only helps us locate a certain element or find the sum up to that element.

Then secondly, I didnt see that $log$ sorry!

We had

$-\ln(1-z) = u'$
$u = -\frac{1}{1-z}$

We have a fraction successfully, all we need now is the $n$.

We can multiply and divide by $n$

$$u = -\frac{2n}{2n - n} = -\frac{2}{1}$$

Never mind, this won't work.

I don't understand the step you made!

Let us focus on generating the $2n-1$ term on the denominator. You can do a trick. For example you can substitute $z^2$ to get

$$\sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$$

Now if we intgrate we get $2n+1$ in the denominator but we want $2n-1$ ! How to get around this ?

 

[Amad27]

Yes , as I said it only helps us locate a certain element or find the sum up to that element.
I don't understand the step you made!

Let us focus on generating the $2n-1$ term on the denominator. You can do a trick. For example you can substitute $z^2$ to get

$$\sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$$

Now if we intgrate we get $2n+1$ in the denominator but we want $2n-1$ ! How to get around this ?

Well, if we differentiate we will get,

$$\frac{-1}{-2z}$$

We can't really integrate

$$-\log(1 -z^2)$$

I mean you can, but it wont help since we will keep on rolling with logs then.Thanks!

 

[alyafey22]

Well, if we differentiate we will get,

$$\frac{-1}{-2z}$$

We can't really integrate

$$-\log(1 -z^2)$$

I mean you can, but it wont help since we will keep on rolling with logs then.Thanks!

You have done the differentiation wrongly. Don't think of the right hand side yet. Let us focus on getting the $2n-1$ term , first.

 

[Amad27]

You have done the differentiation wrongly. Don't think of the right hand side yet. Let us focus on getting the $2n-1$ term , first.

Hi,

We aren't getting anywhere, can we try something new? I mean this:

We are trying to evaluate $\psi(1/2)$

Can you show me a step-by-step evaluation for $\psi(1)$ so that I can see what you are doing and I can then apply that, rather than shooting arrows in the sky.

I honestly feel that is a better strategy once i get an experienced FEEL at it.

What do you say? I think it is better; but ultimately the choice depends on you =)

 

[alyafey22]

Can you show me a step-by-step evaluation for $\psi(1)$

There is no thing to evaluate since we have

$$\psi(z+1) = -\gamma + \sum \frac{z}{n(n+z)}$$

Putting $z=0$ we get $\psi(1) = -\gamma$.

I honestly feel that is a better strategy once i get an experienced FEEL at it.

What do you say? I think it is better; but ultimately the choice depends on you =)

No problem. Ok, we arrived at the following

$$\sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$$

Divide by $z^2$

$$\sum_{n=1}^\infty \frac{z^{2n-2}}{n} = -\frac{\log(1-z^2)}{z^2}$$

Integrate with respect to $z$

$$\sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$$

Put $z=1$

$$\sum_{n=1}^\infty \frac{1}{n(2n-1)} = -\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$Can you evaluate the integral on the right ?

 

[Amad27]

There is no thing to evaluate since we have

$$\psi(z+1) = -\gamma + \sum \frac{z}{n(n+z)}$$

Putting $z=0$ we get $\psi(1) = -\gamma$.
No problem. Ok, we arrived at the following

$$\sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$$

Divide by $z^2$

$$\sum_{n=1}^\infty \frac{z^{2n-2}}{n} = -\frac{\log(1-z^2)}{z^2}$$

Integrate with respect to $z$

$$\sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$$

Put $z=1$

$$\sum_{n=1}^\infty \frac{1}{n(2n-1)} = -\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$Can you evaluate the integral on the right ?

Thank you very much @ZaidAlyafey, kudos to you, and I really appreciate this; the method is much clearer this way as I can understand better, and as I read carefully, all at once. So thank you ZaidAlyafey, I can't thank you enough! I owe you =)

The INTEGRAL:
----------------------
$$-\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$

$$B(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)} \,dt$$

Let's differentiate with respect to $b$

$$B_b(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)}\ln(1-t) \,dt$$

Let $t = x^2 \implies dt = 2x dx$

$$B_b(a,b) = (2) \int_{0}^{1} {x}^{2a-1}{(1-x^2)}^{(b-1)}\ln(1-x^2) \,dx$$

We need $b - 1 = 0 \implies b = 1$
We need $2a - 1 = -2 \implies a = -1/2$

$$B_b((-1/2),1) = (2) \int_{0}^{1} {x}^{-2}\ln(1-x^2) \,dx$$

Darn... We still can't do anything =(

$$B_b(a, b) = B(a, b)(\psi(b) - \psi(a + b))$$

Let $\psi(1/2) = u$

$$I = 1/2B_b(-1/2, 1) = 1/2B(-1/2, 1)(-\gamma - u)$$

The sum is $u = \psi(1/2)$ so

$$u = (-1/2)B(-1/2, 1)\gamma - (1/2)B(-1/2, 1)(u)$$

It is very surprising.

I couldn't compute it but how in the world is $B(-1/2, 1) = -2$? I computed it on WolframAlpha, but I am stumped.

$$u = \gamma + u$$

$$0 = \gamma$$

Some error happened unfortunately.

 

[alyafey22]

Thank you very much @ZaidAlyafey, kudos to you, and I really appreciate this; the method is much clearer this way as I can understand better, and as I read carefully, all at once. So thank you ZaidAlyafey, I can't thank you enough! I owe you =)

The INTEGRAL:
----------------------
$$-\int^1_0\frac{\log(1-x^2)}{x^2}\,dx$$

$$B(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)} \,dt$$

Let's differentiate with respect to $b$

$$B_b(a, b) = \int_{0}^{1} {t}^{a-1}{(1-t)}^{(b-1)}\ln(1-t) \,dt$$

Let $t = x^2 \implies dt = 2x dx$

$$B_b(a,b) = (2) \int_{0}^{1} {x}^{2a-1}{(1-x^2)}^{(b-1)}\ln(1-x^2) \,dx$$

We need $b - 1 = 0 \implies b = 1$
We need $2a - 1 = -2 \implies a = -1/2$

$$B_b((-1/2),1) = (2) \int_{0}^{1} {x}^{-2}\ln(1-x^2) \,dx$$

Darn... We still can't do anything =(

You will run into a cycle. We want to find the integral without using the digamma function. Fortunately , there is an anti-derivative for the integral. Please , give it more thoughts.

The sum is $u = \psi(1/2)$ so

$$u = (-1/2)B(-1/2, 1)\gamma - (1/2)B(-1/2, 1)(u)$$

It is very surprising.

I couldn't compute it but how in the world is $B(-1/2, 1) = -2$? I computed it on WolframAlpha, but I am stumped.

$$u = \gamma + u$$

$$0 = \gamma$$

Some error happened unfortunately.

Actually $I \neq \psi(1/2)$. Remember that we have

$$\psi(1/2) = -\gamma -\sum \frac{1}{n(2n-1)}$$

$$I = -\int^1_0 \frac{\log(1-x^2)}{x^2}\,dx=-\sum \frac{1}{n(2n-1)}= u + \gamma$$

, hence $u+\gamma = u +\gamma$ which holds true. As I said you run into a loop by doing this method.

As how to evaluate $B(-1/2,1)$ by definition we have

$$B(-1/2,1) = \frac{\Gamma(-1/2)\Gamma(1)}{\Gamma(1/2)}$$

It is known that $\Gamma(z+1) = z\Gamma(z) $ so $\Gamma(1/2) = (-1/2)\Gamma(-1/2)$.

 

[Amad27]

You will run into a cycle. We want to find the integral without using the digamma function. Fortunately , there is an anti-derivative for the integral. Please , give it more thoughts.
Actually $I \neq \psi(1/2)$. Remember that we have

$$\psi(1/2) = -\gamma -\sum \frac{1}{n(2n-1)}$$

$$I = -\int^1_0 \frac{\log(1-x^2)}{x^2}\,dx=-\sum \frac{1}{n(2n-1)}= u + \gamma$$

, hence $u+\gamma = u +\gamma$ which holds true. As I said you run into a loop by doing this method.

As how to evaluate $B(-1/2,1)$ by definition we have

$$B(-1/2,1) = \frac{\Gamma(-1/2)\Gamma(1)}{\Gamma(1/2)}$$

It is known that $\Gamma(z+1) = z\Gamma(z) $ so $\Gamma(1/2) = (-1/2)\Gamma(-1/2)$.

Okay.

Do you mind giving a small hint?

So will the integral require gamma or beta?

So I can start it knowing a fact =)

Thanks!

 

[alyafey22]

Okay.

Do you mind giving a small hint?

So will the integral require gamma or beta?

So I can start it knowing a fact =)

Thanks!

It can be solved using elemetnary functions. Moreover , you can find an antiderivative of

$$\int \frac{\log(1-x^2)}{x^2}\,dx$$

Hint: start by $t=1/x^2$.

 

[Amad27]

It can be solved using elemetnary functions. Moreover , you can find an antiderivative of

$$\int \frac{\log(1-x^2)}{x^2}\,dx$$

Hint: start by $t=1/x^2$.

How did you find out that specific substition? It is not obvious to start with $t = 1/x^2$ so how did you figure out that we should make that substitution?

It may seem that I am being pedantic, but knowing the reason behind every step is important.

Sorry if this is too unusual, tell me so. =)

 

[alyafey22]

How did you find out that specific substition? It is not obvious to start with $t = 1/x^2$ so how did you figure out that we should make that substitution?

It may seem that I am being pedantic, but knowing the reason behind every step is important.

Sorry if this is too unusual, tell me so. =)

I was going to sleep then I remembered that I made a type. The substitution was supposed to be $t=1/x$. I am really sorry .

As for the reason for the sub , we have $1/x^2$ which is the derivative multiplied by a constant.

 

[alyafey22]

Ok , here is the how to do it

$$\int \frac{\log(1-x^2)}{x^2}\,dx = -\int \log(1-1/t^2) = -\int\log(t^2-1)-2\log(t)\,dt$$

This can be written as

$$-\int\log(t-1)\,dt -\int \log(t+1)\,dt+2\int\log(t)\,dt $$

All the functions above has an ant-derivative.

 

[Amad27]

Ok , here is the how to do it

$$\int \frac{\log(1-x^2)}{x^2}\,dx = -\int \log(1-1/t^2) = -\int\log(t^2-1)-2\log(t)\,dt$$

This can be written as

$$-\int\log(t-1)\,dt -\int \log(t+1)\,dt+2\int\log(t)\,dt $$

All the functions above has an ant-derivative.

Hello,

Okay let's get

$$-\int \log(t-1)\,dt = -(t-1)\ln(t-1) + (t-1)$$
$$-\int \ln(1 + t) dt = -(1+t)\ln(t+1) + (t+1)$$
$$2\int \ln(t) dt = 2t\ln(t) - 2t $$

$$ \implies -(t-1)\ln(t-1) + (t-1) -(1+t)\ln(t+1) + (t+1) + 2t\ln(t) - 2t$$

How can we change the lower bound in terms of $t$

$$ t = 1/x$$
$$t = 1/0$$

Thanks!

 

[alyafey22]

How can we change the lower bound in terms of $t$

$$ t = 1/x$$
$$t = 1/0$$

Thanks!

We just put $t=1/x$. Try then to find the integral on the interval [0,1].

 

[Amad27]

We just put $t=1/x$. Try then to find the integral on the interval [0,1].

But we CANNOT change the lower bound.

$t = 1/0$ because $x = 0$ was the lower bound.

 

[alyafey22]

But we CANNOT change the lower bound.

$t = 1/0$ because $x = 0$ was the lower bound.

The lower and upper limits before the substitution were $0$ and $1$ consecutively. If we make the substitution they are $\infty$ and $1$. You have the choice between $x$ and $t$. Remember if we have an improper integral we take the limit

$$\int^\infty_1 f(x) \,dx = \lim_{t \to \infty }\int^t_1 f(x) \,dx$$

 

[Amad27]

The lower and upper limits before the substitution were $0$ and $1$ consecutively. If we make the substitution they are $\infty$ and $1$. You have the choice between $x$ and $t$. Remember if we have an improper integral we take the limit

$$\int^\infty_1 f(x) \,dx = \lim_{t \to \infty }\int^t_1 f(x) \,dx$$

Hello,

@ZaidAlyafey, I hope you did not misunderstand that I was trying to ignore you. I did indeed read this reply, but I stood off for a while, getting more in depth with this topic, so now I can discuss this. Please do not mind, I appreciate every bit and I won't give up on this, until I figure it out! =)

I have labeled sections of questions, I would appreciate if you respond to each section separately.

QUESTION #1:
You had this: (We were trying to evaluate $\psi(1)$ as an example)

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$

How did you arrive at this result?

QUESTION #2:
Secondly: You stated the following:

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$

But you did not integrate the LHS. Can you take me behind the scenes here?

QUESTION #3:
Please tell me. Can you write sums as integrals for every sum? If so, how? Thanks!

Final comment:
Again, you did not give up on me, and I did not give up on you either! I just took a break so I could read more (as you have seen from my posts). Please do not mind this, I try not to give up on anything! Thanks for your dedication.

 

[alyafey22]

QUESTION #1:
You had this: (We were trying to evaluate $\psi(1)$ as an example)

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$

How did you arrive at this result?

Ok we start by the following

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

Then integrate to get

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

Finally let $x= z^2$

---

QUESTION #2:
Secondly: You stated the following:

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$

But you did not integrate the LHS. Can you take me behind the scenes here?

From the first part we proved that

$$\sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$$

Now divide by $z^2$ to obtain

$$\sum_{n=1}^\infty \frac{z^{2n-2}}{n} = -\frac{\log(1-z^2)}{z^2}$$

Now integrate with respect to $z$

$$\sum_{n=1}^\infty \int^z_0 \frac{z^{2n-2}}{n} \,dz= -\int^z_0\frac{\log(1-z^2)}{z^2}\,dz$$

---

QUESTION #3:
Please tell me. Can you write sums as integrals for every sum? If so, how? Thanks!

Definite integrals are Riemann sums , so it must be possible but might be complicated. Moreover , it will not necessarily make the problem easier to solve.

---

Final comment:
Again, you did not give up on me, and I did not give up on you either! I just took a break so I could read more (as you have seen from my posts). Please do not mind this, I try not to give up on anything! Thanks for your dedication.

It is good that you took your time trying to understand series.

 

[Amad27]

Ok we start by the following

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

Then integrate to get

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

Finally let $x= z^2$

---

From the first part we proved that

$$\sum_{n=1}^\infty \frac{z^{2n}}{n} = -\log(1-z^2)$$

Now divide by $z^2$ to obtain

$$\sum_{n=1}^\infty \frac{z^{2n-2}}{n} = -\frac{\log(1-z^2)}{z^2}$$

Now integrate with respect to $z$

$$\sum_{n=1}^\infty \int^z_0 \frac{z^{2n-2}}{n} \,dz= -\int^z_0\frac{\log(1-z^2)}{z^2}\,dz$$

---

Definite integrals are Riemann sums , so it must be possible but might be complicated. Moreover , it will not necessarily make the problem easier to solve.

---

It is good that you took your time trying to understand series.

It has been known that:

QUESTION #1:

$\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$

Let $u = n-1$ so

$\displaystyle \sum_{n=1}^{\infty} x^{n-1} = \sum_{u=0}^{\infty} x^{u} = \frac{1}{1-x}$

Is this how you derive that result?

QUESTION #2:
About interchanging summation and integral, doesn't the series have to be a monotone series? How did you test if that series was monotone?
QUESTION #3:

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$

You cannot simply get $\psi(1)$ just be evaluating the integral. The LHS is NOT the proper definition of the digamma $\psi$ function. Wikipedia states:

$\psi(z) = \gamma + \displaystyle \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$

So what next?

Thanks Zaid!

 

[alyafey22]

It has been known that:

QUESTION #1:

$\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$

Let $u = n-1$ so

$\displaystyle \sum_{n=1}^{\infty} x^{n-1} = \sum_{u=0}^{\infty} x^{u} = \frac{1}{1-x}$

Is this how you derive that result?

Yes , exactly.

QUESTION #2:

About interchanging summation and integral, doesn't the series have to be a monotone series? How did you test if that series was monotone?

The function is a geometric series , so you can integrate term by term.

QUESTION #3:

$\displaystyle \sum_{n=1}^\infty \frac{z^{2n-1}}{n(2n-1)} = -\int^z_0\frac{\log(1-x^2)}{x^2}\,dx$

You cannot simply get $\psi(1)$ just be evaluating the integral. The LHS is NOT the proper definition of the digamma $\psi$ function. Wikipedia states:

$\psi(z) = \gamma + \displaystyle \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$

Revise post #23.

 

[Amad27]

Yes , exactly.
The function is a geometric series , so you can integrate term by term.
Revise post #23.

Hello, I just reread that post (long time ago!)

So let $z = 1$ then you have the required sum? That makes sense.

I suppose you have tons of practice with this stuff. I just can't think like that. Perhaps its the lack of my practice (maybe due to my age). Thanks though. Let's try another one?

$\psi(4) = \psi(3 + 1)$, so $z = 3$

$\psi(z+1) = \gamma + \displaystyle \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$

When $z = 3$, $\psi(3+1) = \gamma + \displaystyle 3\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$

$\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$

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

Where $u = n + 2$

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

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

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

$= \displaystyle \sum_{n=0}^{\infty} \frac{(z-3)^{n+3}}{z(n+3)} = \displaystyle \frac{\int_{0}^{z-3} \frac{1}{1-x} - (x + x^2) \,dx}{z}$
Anything that may help?

Thanks!

 

[alyafey22]

We have already proved that

\(\displaystyle \sum_{n=1}^\infty \frac{z^n }{n} = -\log(1-z)\)

Multiply by $z^2$ to get

\(\displaystyle \sum_{n=1}^\infty \frac{z^{n+2} }{n} = -z^2\log(1-z)\)

Now integrate

\(\displaystyle \sum_{n=1}^\infty \frac{z^{n+3} }{n(n+3)} = -\int^z_0 x^2\log(1-x)\,dx\)

Now set $z=1$

\(\displaystyle \sum_{n=1}^\infty \frac{1 }{n(n+3)} = -\int^1_0 x^2\log(1-x)\,dx\)

That should be easy solvable using $t = 1-x$.

Note you can also use

$$\psi(z+1) = \psi(z)+\frac{1}{z}$$.

 

[Amad27]

We have already proved that

\(\displaystyle \sum_{n=1}^\infty \frac{z^n }{n} = -\log(1-z)\)

Multiply by $z^2$ to get

\(\displaystyle \sum_{n=1}^\infty \frac{z^{n+2} }{n} = -z^2\log(1-z)\)

Now integrate

\(\displaystyle \sum_{n=1}^\infty \frac{z^{n+2} }{n(n+3)} = -\int^z_0 x^2\log(1-x)\,dx\)

Now set $z=1$

\(\displaystyle \sum_{n=1}^\infty \frac{1 }{n(n+3)} = -\int^1_0 x^2\log(1-x)\,dx\)

That should be easy solvable using $t = 1-x$.

Note you can also use

$$\psi(z+1) = \psi(z)+\frac{1}{z}$$.

That is fantastic. Let's find

$\psi(5/4) = \psi(1 + 1/4)$ therefore, $z = 1/4$

There is one issue with the integration (not yours but in general).

When you integrate you change the $Z$'s in the integrand to $x$'s. But you do not do the same on the RHS. When you do:

\(\displaystyle \sum_{n=1}^\infty \frac{z^{n+2} }{n} = -z^2\log(1-z)\)

Now integrate

\(\displaystyle \sum_{n=1}^\infty \frac{z^{n+2} }{n(n+3)} = -\int^z_0 x^2\log(1-x)\,dx\)

How?

 

[alyafey22]

That is fantastic. Let's find

$\psi(5/4) = \psi(1 + 1/4)$ therefore, $z = 1/4$

Let us see what you have tried.

When you integrate you change the $Z$'s in the integrand to $x$'s. But you do not do the same on the RHS. When you do:

How?

I made a mistake it should be

\(\displaystyle \sum_{n=1}^\infty\frac{z^{n+3}}{n(n+3)} =- \int^z_0 x^2\log(1-x)\,dx\)

Since it is a dummy variable you can use $z$ but I am trying to avoid confusion. The following is also correct

\(\displaystyle \sum_{n=1}^\infty\frac{z^{n+3}}{n(n+3)} = -\int^z_0 z^2\log(1-z)\,dz\)

 

[Amad27]

Let us see what you have tried.
I made a mistake it should be

\(\displaystyle \sum_{n=1}^\infty\frac{z^{n+3}}{n(n+3)} =- \int^z_0 x^2\log(1-x)\,dx\)

Since it is a dummy variable you can use $z$ but I am trying to avoid confusion. The following is also correct

\(\displaystyle \sum_{n=1}^\infty\frac{z^{n+3}}{n(n+3)} = -\int^z_0 z^2\log(1-z)\,dz\)

So then,

Let $z = 1$ On the RHS you have,

\(\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+3)} = -\int^1_0 \log(1-1)\,dz\)

which is not possible.

 

[alyafey22]

So then,

Let $z = 1$ On the RHS you have,

\(\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+3)} = -\int^1_0 \log(1-1)\,dz\)

which is not possible.

This is where a confusion happens. The inner $z$ is independent of the outer $z$ so to avoid that we always avoid such symbols. The inner $z$ is a dummy variable so we can choose any thing we want $x$ , $y$ or even $z$. So it is always preferable to use a symbol different than the boundary of integration for example $x$

$$\int^z_0 x^2\log(1-x)\,dx$$