How Can the Given Definite Integral Identity Be Proven for Any Natural Number n?

[lfdahl]

Show, that the identity

\[\int_{0}^{1}\frac{x^{n-1}+x^{n-\frac{1}{2}}-2x^{2n-1}}{1-x}dx = 2\ln2\]

- holds for any natural number $n$.

 

[lfdahl]

Here´s the suggested solution:

Spoiler

Rewriting the integrand:

\[\frac{x^{n-1}+x^{n-1/2}-2x^{2n-1}}{1-x}=\frac{x^{n-1}(1+\sqrt{x}-2x^n)}{1-x}=\frac{x^{n-1}((1-\sqrt{x})(1+2\sqrt{x})+2x-2x^n)}{1-x}\]

\[=\frac{x^{n-1}(1+2\sqrt{x})}{1+\sqrt{x}}+\frac{2x^n(1-x^{n-1})}{1-x}\\\\=x^{n-1}+x^{n-1}\frac{\sqrt{x}}{1+\sqrt{x}}+2x^n\:\frac{(1-x)(1+x+...+x^{n-3}+x^{n-2})}{1-x} \\\\=2x^{n-1}-\frac{x^{n-1}}{1+\sqrt{x}}+2\left ( x^n+x^{n+1}+...+x^{2n-2} \right ) \\\\=2\left ( x^{n-1}+x^n+...+x^{2n-2} \right )-\frac{x^{n-1}}{1+\sqrt{x}}\]

Integrating yields:
\[I = \int_{0}^{1} \left ( 2\left ( x^{n-1}+x^n+...+x^{2n-2} \right )-\frac{x^{n-1}}{1+\sqrt{x}} \right )dx = 2\sum_{j=n}^{2n-1}\frac{1}{j}-\int_{0}^{1}\frac{x^{n-1}}{1+\sqrt{x}}dx\]
The last term can be rewritten as ($x = u^2$):
\[\int_{0}^{1}\frac{x^{n-1}}{1+\sqrt{x}}dx = 2\int_{0}^{1}\frac{u^{2n-1}}{1+u }du\]

Rewriting the integrand using the geometric series:

\[\frac{u^{2n-1}}{1+u} = u^{2n-1}\sum_{i=0}^{\infty}(-u)^i = u^{2n-1}-u^{2n}+u^{2n+1}-... \]

The right hand side is thus obtained by cutting off the first $2n-1$ terms of an alternating geometric series:

\[\frac{1}{1+u}=1-u+u^2-u^3+...+u^{2n-2}-u^{2n-1}+u^{2n}-u^{2n+1}... \\\\=1-u+u^2-u^3+...+u^{2n-2}- \left ( u^{2n-1}-u^{2n}+u^{2n+1}... \right )\]

Thus, the integral can be written:

\[2\int_{0}^{1}\frac{u^{2n-1}}{1+u }du = 2\int_{0}^{1}\left ( 1-u+u^2-u^3+...+u^{2n-2}-\frac{1}{1+u} \right )du \\\\=2\left ( \sum_{j=1}^{2n-1}\frac{(-1)^{j-1}}{j}-\ln 2 \right )\]

Our total integral now has the form:

\[I =2\sum_{j=n}^{2n-1}\frac{1}{j}- 2\left ( \sum_{j=1}^{2n-1}\frac{(-1)^{j-1}}{j}-\ln 2 \right )\\\\ =2\sum_{j=n}^{2n-1}\frac{1}{j}-2\left ( \sum_{j=1}^{2n-1}\frac{1}{j} - 2\sum_{j=1}^{n-1}\frac{1}{2j}\right )+2\ln 2 \\\\=2\left ( \sum_{j=1}^{2n-1}\frac{1}{j}-\left ( \sum_{j=1}^{2n-1}\frac{1}{j}\right ) \right )+2\ln 2 \\\\= 2 \ln 2.\]
- and we´re done.

 

[Theia]

I'm sorry about poking a bit older threat, but I found this interesting, so here is my attempt:

Spoiler

Let's define

\(\displaystyle F(n) = \int_0^1 \frac{x^{n-1} + x^{n-1/2} - 2x^{2n-1}}{1-x}\mathrm{d}x \qquad \forall \ n \in \mathbb{N}.\)

One wants to show that the value of \(\displaystyle F(n)\) doesn't change, i.e. it's constant. To show this, one needs to show that \(\displaystyle F'(n) = 0\), where \(\displaystyle '\) denotes the derivative with respect to \(\displaystyle n\).

Let's calculate the derivative by applying Leibniz's rule of derivating under integral sign. One obtains

\(\displaystyle \begin{align*}F'(n) &= \frac{\mathrm{d}F}{\mathrm{d}n} = \int_0^1 \frac{\partial}{\partial n} \left( \frac{x^{n-1} + x^{n-1/2} - 2x^{2n-1}}{1-x} \right) \mathrm{d}x \\
& = \int_0^1 \frac{x^{n-1} + x^{n-1/2} - 4x^{2n-1}}{1-x} \ln x \ \mathrm{d}x \ .\end{align*}\)

Next one uses the geometric serie to rewrite denominator (\(\displaystyle 0 \le x < 1\)):

\(\displaystyle \frac{1}{1-x} = \sum_{v = 0}^{\infty} x^v = 1 + x + x^2 + \cdots . \)

Using this, the derivative of \(\displaystyle F\) can be written as

\(\displaystyle \begin{align*}F'(n) &= \int_0^1 \left( x^{n-1} + x^{n-1/2} - 2x^{2n-1} \right) \ln x \sum_{v = 0}^{\infty} x^v \ \mathrm{d}x \\
&= \int_0^1 \sum_{v = 0}^{\infty} x^{v+n-1} \ln x \ \mathrm{d}x + \int_0^1 \sum_{v = 0}^{\infty} x^{v+n-1/2} \ln x \ \mathrm{d}x \\
&\phantom{=-} -4 \int_0^1 \sum_{v = 0}^{\infty} x^{v+2n-1} \ln x \ \mathrm{d}x \ .\end{align*}\)

Now one can change the order of summation and integration (by e.g. Fubini's theorem) and one obtains

\(\displaystyle \begin{align*}F'(n) &= \sum_{v = 0}^{\infty} \int_0^1 x^{v+n-1} \ln x \ \mathrm{d}x + \sum_{v = 0}^{\infty} \int_0^1 x^{v+n-1/2} \ln x \ \mathrm{d}x \\
&\phantom{=} -4 \sum_{v = 0}^{\infty} \int_0^1 x^{v+2n-1} \ln x \ \mathrm{d}x \ .\end{align*}\)

Now integrations are trivial and one can write

\(\displaystyle \begin{align*}F'(n) &= \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+n-1} \ln x + \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+n-1/2} \ln x \\
&\phantom{=} -4 \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+2n-1} \ln x \\
&= -\sum_{v=0}^{\infty} \frac{1}{\left( v+n \right) ^2} - \sum_{v=0}^{\infty} \frac{1}{\left( v+n+\tfrac{1}{2} \right) ^2} + 4 \sum_{v=0}^{\infty} \frac{1}{\left( v+2n \right) ^2} . \end{align*}\)

In terms of polygamma function \(\displaystyle \psi ^{(m)}\) this can be written

\(\displaystyle F'(n) = 4\psi ^{(1)}(2n) - \psi ^{(1)}(n + \tfrac{1}{2}) - \psi ^{(1)}(n) \ . \)

By using the multiplication formula one can rewrite the \(\displaystyle 4\psi ^{(1)}(2n)\):

\(\displaystyle 4\psi ^{(1)}(2n) = \psi ^{(1)}(n + \tfrac{1}{2}) + \psi ^{(1)}(n) \ , \)

which gives the result \(\displaystyle F'(n) = 0 \ n \in \mathbb{N}\), and hence \(\displaystyle F(n) = \textrm{constant}.\)

The value of \(\displaystyle F(n)\) one can evaluate by substituting e.g. \(\displaystyle n = 1\), which gives

\(\displaystyle \begin{align*} F(n) &= F(1) = \int_0^1 \frac{x^0 + x^{1/2} - 2x^1}{1-x}\mathrm{d}x \\
&= {\Big/}_{\!\!\!0}^{1} \left( 2x - 2\sqrt{x} + 2 \ln \left( \sqrt{x} + 1 \right) \right) \\
&= 2 \ln 2 , \end{align*}\)

which is what one wanted to show. QED.

 

[lfdahl]

Thankyou, Theia, for a clever solution!

 

[topsquark]

I'm sorry about poking a bit older threat, but I found this interesting, so here is my attempt:

Spoiler

Let's define

\(\displaystyle F(n) = \int_0^1 \frac{x^{n-1} + x^{n-1/2} - 2x^{2n-1}}{1-x}\mathrm{d}x \qquad \forall \ n \in \mathbb{N}.\)

One wants to show that the value of \(\displaystyle F(n)\) doesn't change, i.e. it's constant. To show this, one needs to show that \(\displaystyle F'(n) = 0\), where \(\displaystyle '\) denotes the derivative with respect to \(\displaystyle n\).

Let's calculate the derivative by applying Leibniz's rule of derivating under integral sign. One obtains

\(\displaystyle \begin{align*}F'(n) &= \frac{\mathrm{d}F}{\mathrm{d}n} = \int_0^1 \frac{\partial}{\partial n} \left( \frac{x^{n-1} + x^{n-1/2} - 2x^{2n-1}}{1-x} \right) \mathrm{d}x \\
& = \int_0^1 \frac{x^{n-1} + x^{n-1/2} - 4x^{2n-1}}{1-x} \ln x \ \mathrm{d}x \ .\end{align*}\)

Next one uses the geometric serie to rewrite denominator (\(\displaystyle 0 \le x < 1\)):

\(\displaystyle \frac{1}{1-x} = \sum_{v = 0}^{\infty} x^v = 1 + x + x^2 + \cdots . \)

Using this, the derivative of \(\displaystyle F\) can be written as

\(\displaystyle \begin{align*}F'(n) &= \int_0^1 \left( x^{n-1} + x^{n-1/2} - 2x^{2n-1} \right) \ln x \sum_{v = 0}^{\infty} x^v \ \mathrm{d}x \\
&= \int_0^1 \sum_{v = 0}^{\infty} x^{v+n-1} \ln x \ \mathrm{d}x + \int_0^1 \sum_{v = 0}^{\infty} x^{v+n-1/2} \ln x \ \mathrm{d}x \\
&\phantom{=-} -4 \int_0^1 \sum_{v = 0}^{\infty} x^{v+2n-1} \ln x \ \mathrm{d}x \ .\end{align*}\)

Now one can change the order of summation and integration (by e.g. Fubini's theorem) and one obtains

\(\displaystyle \begin{align*}F'(n) &= \sum_{v = 0}^{\infty} \int_0^1 x^{v+n-1} \ln x \ \mathrm{d}x + \sum_{v = 0}^{\infty} \int_0^1 x^{v+n-1/2} \ln x \ \mathrm{d}x \\
&\phantom{=} -4 \sum_{v = 0}^{\infty} \int_0^1 x^{v+2n-1} \ln x \ \mathrm{d}x \ .\end{align*}\)

Now integrations are trivial and one can write

\(\displaystyle \begin{align*}F'(n) &= \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+n-1} \ln x + \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+n-1/2} \ln x \\
&\phantom{=} -4 \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+2n-1} \ln x \\
&= -\sum_{v=0}^{\infty} \frac{1}{\left( v+n \right) ^2} - \sum_{v=0}^{\infty} \frac{1}{\left( v+n+\tfrac{1}{2} \right) ^2} + 4 \sum_{v=0}^{\infty} \frac{1}{\left( v+2n \right) ^2} . \end{align*}\)

In terms of polygamma function \(\displaystyle \psi ^{(m)}\) this can be written

\(\displaystyle F'(n) = 4\psi ^{(1)}(2n) - \psi ^{(1)}(n + \tfrac{1}{2}) - \psi ^{(1)}(n) \ . \)

By using the multiplication formula one can rewrite the \(\displaystyle 4\psi ^{(1)}(2n)\):

\(\displaystyle 4\psi ^{(1)}(2n) = \psi ^{(1)}(n + \tfrac{1}{2}) + \psi ^{(1)}(n) \ , \)

which gives the result \(\displaystyle F'(n) = 0 \ n \in \mathbb{N}\), and hence \(\displaystyle F(n) = \textrm{constant}.\)

The value of \(\displaystyle F(n)\) one can evaluate by substituting e.g. \(\displaystyle n = 1\), which gives

\(\displaystyle \begin{align*} F(n) &= F(1) = \int_0^1 \frac{x^0 + x^{1/2} - 2x^1}{1-x}\mathrm{d}x \\
&= {\Big/}_{\!\!\!0}^{1} \left( 2x - 2\sqrt{x} + 2 \ln \left( \sqrt{x} + 1 \right) \right) \\
&= 2 \ln 2 , \end{align*}\)

which is what one wanted to show. QED.

Interesting. I note that Theia's solution uses a derivative of n, which implies that n doesn't have to be natural number! (I double-checked this on W|A.)

-Dan

 

[Theia]

Yes, \(\displaystyle n \in \mathbb{N}\) is not a crucial requirement for \(\displaystyle F'(n) = 0\). But I haven't yet found a proof that shows what is required for \(\displaystyle n\).(Wondering)

 

[MountEvariste]

I'm sorry about poking a bit older threat, but I found this interesting, so here is my attempt:

Spoiler

Let's define

\(\displaystyle F(n) = \int_0^1 \frac{x^{n-1} + x^{n-1/2} - 2x^{2n-1}}{1-x}\mathrm{d}x \qquad \forall \ n \in \mathbb{N}.\)

One wants to show that the value of \(\displaystyle F(n)\) doesn't change, i.e. it's constant. To show this, one needs to show that \(\displaystyle F'(n) = 0\), where \(\displaystyle '\) denotes the derivative with respect to \(\displaystyle n\).

Let's calculate the derivative by applying Leibniz's rule of derivating under integral sign. One obtains

\(\displaystyle \begin{align*}F'(n) &= \frac{\mathrm{d}F}{\mathrm{d}n} = \int_0^1 \frac{\partial}{\partial n} \left( \frac{x^{n-1} + x^{n-1/2} - 2x^{2n-1}}{1-x} \right) \mathrm{d}x \\
& = \int_0^1 \frac{x^{n-1} + x^{n-1/2} - 4x^{2n-1}}{1-x} \ln x \ \mathrm{d}x \ .\end{align*}\)

Next one uses the geometric serie to rewrite denominator (\(\displaystyle 0 \le x < 1\)):

\(\displaystyle \frac{1}{1-x} = \sum_{v = 0}^{\infty} x^v = 1 + x + x^2 + \cdots . \)

Using this, the derivative of \(\displaystyle F\) can be written as

\(\displaystyle \begin{align*}F'(n) &= \int_0^1 \left( x^{n-1} + x^{n-1/2} - 2x^{2n-1} \right) \ln x \sum_{v = 0}^{\infty} x^v \ \mathrm{d}x \\
&= \int_0^1 \sum_{v = 0}^{\infty} x^{v+n-1} \ln x \ \mathrm{d}x + \int_0^1 \sum_{v = 0}^{\infty} x^{v+n-1/2} \ln x \ \mathrm{d}x \\
&\phantom{=-} -4 \int_0^1 \sum_{v = 0}^{\infty} x^{v+2n-1} \ln x \ \mathrm{d}x \ .\end{align*}\)

Now one can change the order of summation and integration (by e.g. Fubini's theorem) and one obtains

\(\displaystyle \begin{align*}F'(n) &= \sum_{v = 0}^{\infty} \int_0^1 x^{v+n-1} \ln x \ \mathrm{d}x + \sum_{v = 0}^{\infty} \int_0^1 x^{v+n-1/2} \ln x \ \mathrm{d}x \\
&\phantom{=} -4 \sum_{v = 0}^{\infty} \int_0^1 x^{v+2n-1} \ln x \ \mathrm{d}x \ .\end{align*}\)

Now integrations are trivial and one can write

\(\displaystyle \begin{align*}F'(n) &= \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+n-1} \ln x + \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+n-1/2} \ln x \\
&\phantom{=} -4 \sum_{v = 0}^{\infty} {\Big/}_{\!\!\!0}^{1} x^{v+2n-1} \ln x \\
&= -\sum_{v=0}^{\infty} \frac{1}{\left( v+n \right) ^2} - \sum_{v=0}^{\infty} \frac{1}{\left( v+n+\tfrac{1}{2} \right) ^2} + 4 \sum_{v=0}^{\infty} \frac{1}{\left( v+2n \right) ^2} . \end{align*}\)

In terms of polygamma function \(\displaystyle \psi ^{(m)}\) this can be written

\(\displaystyle F'(n) = 4\psi ^{(1)}(2n) - \psi ^{(1)}(n + \tfrac{1}{2}) - \psi ^{(1)}(n) \ . \)

By using the multiplication formula one can rewrite the \(\displaystyle 4\psi ^{(1)}(2n)\):

\(\displaystyle 4\psi ^{(1)}(2n) = \psi ^{(1)}(n + \tfrac{1}{2}) + \psi ^{(1)}(n) \ , \)

which gives the result \(\displaystyle F'(n) = 0 \ n \in \mathbb{N}\), and hence \(\displaystyle F(n) = \textrm{constant}.\)

The value of \(\displaystyle F(n)\) one can evaluate by substituting e.g. \(\displaystyle n = 1\), which gives

\(\displaystyle \begin{align*} F(n) &= F(1) = \int_0^1 \frac{x^0 + x^{1/2} - 2x^1}{1-x}\mathrm{d}x \\
&= {\Big/}_{\!\!\!0}^{1} \left( 2x - 2\sqrt{x} + 2 \ln \left( \sqrt{x} + 1 \right) \right) \\
&= 2 \ln 2 , \end{align*}\)

which is what one wanted to show. QED.

It really is an interesting question.

I've been thinking about it this evening, and wonder if what's below makes sense.

Spoiler

I was thinking about an alternative way to show that the derivative vanishes in your solution. For $\alpha \ge 1$, let $\displaystyle f(\alpha) = \int_0^1 \frac{x^{\alpha-1}+x^{\alpha-\frac{1}{2}}-2x^{2\alpha-1}}{1-x}\,\mathrm{dx}.$ Then $\displaystyle f'(\alpha) = \int_0^1 \frac{x^{\alpha-1}+x^{\alpha-\frac{1}{2}}-4x^{2\alpha-1}}{1-x} \log(x)\,\mathrm{dx}.$

For any $x \in (0, 1)$ we have $(1-\frac{1}{x}) \leqslant \log(x) \leqslant x-1. $ Applying this to the integral, we have

$\displaystyle \int_0^1 \frac{x^{\alpha-1}+x^{\alpha-\frac{1}{2}}-4x^{2\alpha-1}}{1-x} (1-\frac{1}{x})\,\mathrm{dx} \leqslant f'(\alpha) \leqslant \int_0^1 \frac{(x^{\alpha-1}+x^{\alpha-\frac{1}{2}}-4x^{2\alpha-1}(x-1)}{1-x} \,\mathrm{dx}$

$\implies \displaystyle \int_0^1 {4x^{2\alpha-2}-x^{\alpha-\frac{3}{2}}-x^{\alpha-2}
} \,\mathrm{dx} \leqslant f'(\alpha)\leqslant \int_0^1(4x^{2\alpha-1} -x^{\alpha-\frac{1}{2}}-x^{\alpha-1}) \,\mathrm{dx} $

Hence $ \displaystyle \frac{1}{-2\alpha^2+3\alpha-1} \leqslant f'(\alpha) \leqslant \frac{1}{2\alpha^2+\alpha}$. Solving these with the assumption $\alpha \geqslant 1$ we've:

$ \displaystyle \frac{1}{-2\alpha^2+3\alpha-1} \leqslant f'(\alpha) \implies f'(\alpha) \in [0,\infty); $ and $ \displaystyle f'(\alpha) \leqslant \frac{1}{2\alpha^2+\alpha} \implies f'(\alpha) \in (-\infty,0].$

So $f'(\alpha) \in (-\infty, 0] \cap [0, \infty) = \left\{0\right\}$. So $f'(\alpha) \in \left\{0\right\}$ so $f'(\alpha) = 0$ for all $\alpha \ge 1$. Thus $f$ is constant.

Not sure if I've made a mistake somewhere. But the approach should work nonetheless I feel.

 

[Theia]

Spoiler

Well... It has been a while, isn't it? (Wave)

Getting back to this problem, as it was very interesting. Now I have following solution for it:

Let's define

\[ g(a) = \int_{0} ^{1} \frac{x^{a}}{1 - x} \mathrm{d}x = \sum_{v=0} ^{\infty} \int_{0} ^{1} x^{a+v}\mathrm{d}x = \sum_{v=0} ^{\infty} \frac{1}{a+v+1}. \]

Now the integral in question can be written as

\[ F(n) = g(n-1) + g \left( n-\frac{1}{2} \right) - 2g(2n-1) = \sum_{v=0} ^{\infty} \left( \frac{1}{v + n} + \frac{1}{v + \left( n + \frac{1}{2} \right)} - \frac{2}{v + 2n} \right). \]

By noticing that \( a + v \ne -1 \) one arrives to a condition \( n \ne 0, -\frac{1}{2}, -1, -\frac{3}{2}, \ldots \).

Then by using identities of digamma function one directly obtains

\[ F(n) = 2\psi (2n+1) - \psi (n+1) - \psi \left(n + \frac{3}{2}\right) + \frac{1}{n} + \frac{1}{n + \frac{1}{2}} - \frac{2}{2n} = 2\ln 2.\]

Ok, but what if \( n = 0, -\frac{1}{2} , -1, -\frac{3}{2}, \ldots \)?

Let's write \( n = -\frac{p}{2} \) where \(p = 0, 1, \ldots \). After that one can substitute \( x = \frac{1}{q^{2}} \) and one obtains

\[ F\left( n = -\frac{p}{2} \right) = 2\int_{1} ^{\infty} \frac{q^{p} + q^{p+1} - 2q^{2p+1}}{q^{2} - 1} \mathrm{d}q \].

This is clearly divergent.

Hence it looks like the value of the integral equals to \( 2\ln 2\) as long as \(n \ne 0, -\frac{1}{2}, -1, -\frac{3}{2}, \ldots \).

Any thoughts on this?

 

[lfdahl]

Hello,Theia

Spoiler

Unfortunately, I am not familiar with the digamma function, so I am not competent
to answer and comment on your interesting solution in a proper manner. :unsure:

I do hope, that someone else in this forum can be of help here ...