Prove some identities!

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• Nov 9th 2010, 11:19 AM
Bruno J.
Prove some identities!
There used to be a time when mathematicians were champions of series, infinite products and special functions. If you couldn't give the series expansion of Jacobi's elliptic functions in a minute, there was no way you'd even be admitted to study at Oxford in 1900.

With this said, today most mathematicians don't care so much about series and special functions. This isn't to say that such things are less important, but they are less important in comparison to more modern mathematics. We still very much enjoy seeing the solution to a nice series, because it is pleasing to the eye and to the mind.

In this thread, I will post some kind of series or part of an identity, which has to be evaluated. Feel free to post some as well!

Let's begin with something not too hard. First, evaluate

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

Then evaluate

$\sum_{k=1}^\infty \frac{1}{2^kk^2}$
• Nov 9th 2010, 01:08 PM
Unbeatable0
Good idea! I'm interested to know where this thread will get.

Spoiler:

$
\displaystyle{\sum_{n=1}^\infty}\frac{1}{2^nn} = \sum_{n=1}^\infty}\frac{x^n}{n} \big|_{x=\frac{1}{2}} = -\ln(1-x)\big|_{x=\frac{1}{2}} = \ln 2}
$

Now let $f(x) = \displaystyle{\sum_{n=1}^\infty}\frac{x^n}{n^2}$

Then $\displaystyle{\sum_{n=1}^\infty}\frac{1}{2^nn^2} = f\left(\frac{1}{2}\right)$

$\displaystyle{
f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n} = \frac{1}{x}\sum_{n=1}^\infty \frac{x^n}{n} = \frac{-\ln(1-x)}{x}
}$

Integrate along $[0,x]$ ( $x\in[0,1)$. Note that $f(0)=0$):

$\displaystyle{
f(x) = -\int_0^x \frac{\ln(1-t)}{t} dt = -\int_{1-x}^1 \frac{\ln u}{1-u} du
}$

$\displaystyle{
= -\sum_{n=0}^\infty \int_{1-x}^1 u^n\ln u du = -\sum_{n=0}^\infty \big[\frac{u^{n+1}\ln u}{n+1}-\frac{u^{n+1}}{(n+1)^2}\big]^{1}_{1-x}
}$

$\displaystyle{
= \sum_{n=1}^\infty \left(\frac{(1-x)^n\ln (1-x)}{n}-\frac{(1-x)^n}{n^2}-\frac{1}{n^2}\right)
}$

where we've done integration by substitution and by parts

$\displaystyle{
= \ln (1-x)\sum_{n=1}^\infty \frac{(1-x)^n}{n}-\sum_{n=1}^\infty\frac{(1-x)^n}{n^2} + \sum_{n=1}^\infty \frac{1}{n^2}= -\ln(1-x)\ln x - f(1-x) - \frac{\pi^2}{6}
}$

Plugging $x=\frac{1}{2}$ and solving for $f\left(\frac{1}{2}\right)$ we get

$\displaystyle{\sum_{n=1}^\infty}\frac{1}{2^nn^2} = \frac{1}{12}(\pi^2-6\ln^2 2)$
• Nov 9th 2010, 01:09 PM
chisigma
Is...

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

... so that setting in (1) $x=\frac{1}{2}$ we obtain...

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{2^{k}\ k} = -\ln (1-\frac{1}{2}) = \ln 2$ (2)

Kind regards

$\chi$ $\sigma$
• Nov 9th 2010, 01:42 PM
Bruno J.
Very nice!

Here is another series, not too difficult but still a good challenge. Evaluate

$\displaystyle\sum_{n=1}^\infty \frac{1}{(4n-1)(4n+1)}$

Here is a pretty tough integral.

$\displaystyle\int_0^\infty \frac{x}{\sinh x}dx$
• Nov 9th 2010, 02:30 PM
Also sprach Zarathustra
For the first infinite sum...

Maybe we shall look at d{arcotanh(x)}/dx=1/(1-x^2)... ?
• Nov 9th 2010, 07:27 PM
melese
Quote:

Originally Posted by Bruno J.
Very nice!

Here is another series, not too difficult but still a good challenge. Evaluate

$\displaystyle\sum_{n=1}^\infty \frac{1}{(4n-1)(4n+1)}$

Here is a pretty tough integral.

$\displaystyle\int_0^\infty \frac{x}{\sinh x}dx$

$\displaystyle\sum_{n=1}^\infty \frac{1}{(4n-1)(4n+1)}=\displaystyle\sum_{n=1}^\infty \frac{1}{2}(\frac{1}{4n-1}-\frac{1}{4n+1})=\displaystyle\frac{1}{2}\sum_{n=1} ^\infty (\frac{1}{4n-1}-\frac{1}{4n+1})=\displaystyle\frac{1}{2}\sum_{n=1} ^\infty (\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\cdots)$.

Using the Gregory–Leibniz series $\displaystyle\sum_{n=1}^\infty (1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots)=\pi/4$ we find that the sum of the original sum is: $\frac{1}{2}(1-\frac{\pi}{4})$
• Nov 9th 2010, 08:53 PM
Drexel28
Quote:

Originally Posted by Bruno J.
Here is a pretty tough integral.

$\displaystyle\int_0^\infty \frac{x}{\sinh x}dx$

Spoiler:

Let's do this in more general terms. In other words, let's solve

$\displaystyle I= \int_0^{\infty}\frac{x^{\alpha}}{\text{sinh}(x)}dx \quad\alpha>0$

To do this we merely note that

\displaystyle \begin{aligned}\int_0^{\infty}\frac{x^{\alpha}}{\t ext{sinh}(x)}dx & = \int_0^{\infty}\frac{x^{\alpha}}{\frac{e^x-e^{-x}}{2}}dx\\ &= 2\int_0^{\infty}\frac{e^{-x}x^{\alpha}}{1-\left(e^{-x}\right)^2}dx\end{aligned}

Note thought that since $e^{-2x}\leqslant 1$ we may apply the geometric series to arrive at

\displaystyle \begin{aligned}I &= 2\int_0^{\infty}\frac{e^{-x}x^{\alpha}}{1-e^{-2x}}dx\\ &= 2\int_0^{\infty}e^{-x}x^{\alpha}\sum_{n=0}^{\infty}e^{-2nx}dx\\ &= 2\sum_{n=0}^{\infty}\int_0^{\infty}x^{\alpha}e^{-(2n+1)x}dx\end{aligned}

But, making the substitution $(2n+1)x=z$ gives us

\begin{aligned}I &= 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha+1}}\in t_0^{\infty}z^{\alpha}e^{-\alpha}dz\\ &=2\Gamma(\alpha+1)\sum_{n=0}^{\infty}\frac{1}{(2n +1)^{\alpha+1}}\end{aligned}

Now, to compute this last sum we note that

\displaystyle \begin{aligned}\zeta(\alpha) &= \sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}\\ &=\sum_{n=1}^{\infty}\frac{1}{(2n)^{\alpha}}+\sum_ {n=0}^{\infty}\frac{1}{(2n+1)^{\alpha}}\\ &= \frac{\zeta(\alpha)}{2^{\alpha}}+\sum_{n=0}^{\inft y}\frac{1}{(2n+1)^{\alpha}}\end{aligned}

And thus

$\displaystyle \sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha}} = \zeta(\alpha)\left(1-\frac{1}{2^\alpha}\right)$

Thus, applying this we may finally conclude that

\displaystyle \begin{aligned}I &= 2\Gamma(\alpha+1)\sum_{n=0}^{\infty}\frac{1}{(2n+1 )^{\alpha+1}}\\ &= 2\Gamma(\alpha+1)\zeta(\alpha+1)\left(1-\frac{1}{2^{\alpha+1}}\right)\\ &= \Gamma(\alpha+1)\zeta(\alpha+1)\left(2-\frac{1}{2^{\alpha}}\right)\end{aligned}

• Nov 9th 2010, 08:57 PM
simplependulum
Quote:

Originally Posted by Bruno J.
$\sum_{k=1}^\infty \frac{1}{2^kk^2}$

I think most people will do this as the first step :

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

$\displaystyle = - \int_0^{1/2} \frac{ \ln(1-x) }{x}~dx$

Then , from $\displaystyle \frac{1}{x} + \frac{1}{1-x} = \frac{1}{x(1-x)}$

we have

$\displaystyle S = \int_0^{1/2} \frac{\ln(1-x)}{1-x}~dx - \int_0^{1/2} \frac{\ln(1-x)}{x(1-x)}~dx$

$\displaystyle = \left[-\frac{\ln^2(1-x)}{2} \right]_0^{1/2} - \int_0^{1/2} \frac{\ln(1-x)}{x(1-x)}~dx$

For the second integral , sub $\displaystyle x = \frac{t}{t+1}$ we find that

$\displaystyle \int_0^{1/2} \frac{\ln(1-x)}{x(1-x)}~dx = \int_0^1 \frac{-\ln(t+1)}{\frac{t}{(t+1)^2} }\cdot \frac{dt}{(t+1)^2}$

$\displaystyle = - \int_0^1 \frac{\ln(t+1)}{t}~dt$

Recall another famous integral expression for $\displaystyle \frac{\pi^2}{12} = \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - ... = \int_0^1 \frac{\ln(t+1)}{t}~dt$

Therefore , $\displaystyle S = \frac{\pi^2}{12} - \frac{\ln^2{2}}{2}$
• Nov 10th 2010, 12:12 AM
simplependulum
Quote:

Originally Posted by Drexel28
Spoiler:

Let's do this in more general terms. In other words, let's solve

$\displaystyle I= \int_0^{\infty}\frac{x^{\alpha}}{\text{sinh}(x)}dx \quad\alpha>0$

To do this we merely note that

\displaystyle \begin{aligned}\int_0^{\infty}\frac{x^{\alpha}}{\t ext{sinh}(x)}dx & = \int_0^{\infty}\frac{x^{\alpha}}{\frac{e^x-e^{-x}}{2}}dx\\ &= 2\int_0^{\infty}\frac{e^{-x}x^{\alpha}}{1-\left(e^{-x}\right)^2}dx\end{aligned}

Note thought that since $e^{-2x}\leqslant 1$ we may apply the geometric series to arrive at

\displaystyle \begin{aligned}I &= 2\int_0^{\infty}\frac{e^{-x}x^{\alpha}}{1-e^{-2x}}dx\\ &= 2\int_0^{\infty}e^{-x}x^{\alpha}\sum_{n=0}^{\infty}e^{-2nx}dx\\ &= 2\sum_{n=0}^{\infty}\int_0^{\infty}x^{\alpha}e^{-(2n+1)x}dx\end{aligned}

But, making the substitution $(2n+1)x=z$ gives us

\begin{aligned}I &= 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha+1}}\in t_0^{\infty}z^{\alpha}e^{-\alpha}dz\\ &=2\Gamma(\alpha+1)\sum_{n=0}^{\infty}\frac{1}{(2n +1)^{\alpha+1}}\end{aligned}

Now, to compute this last sum we note that

\displaystyle \begin{aligned}\zeta(\alpha) &= \sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}\\ &=\sum_{n=1}^{\infty}\frac{1}{(2n)^{\alpha}}+\sum_ {n=0}^{\infty}\frac{1}{(2n+1)^{\alpha}}\\ &= \frac{\zeta(\alpha)}{2^{\alpha}}+\sum_{n=0}^{\inft y}\frac{1}{(2n+1)^{\alpha}}\end{aligned}

And thus

$\displaystyle \sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha}} = \zeta(\alpha)\left(1-\frac{1}{2^\alpha}\right)$

Thus, applying this we may finally conclude that

\displaystyle \begin{aligned}I &= 2\Gamma(\alpha+1)\sum_{n=0}^{\infty}\frac{1}{(2n+1 )^{\alpha+1}}\\ &= 2\Gamma(\alpha+1)\zeta(\alpha+1)\left(1-\frac{1}{2^{\alpha+1}}\right)\\ &= \Gamma(\alpha+1)\zeta(\alpha+1)\left(2-\frac{1}{2^{\alpha}}\right)\end{aligned}

So the answer to the original problem is $\displaystyle \frac{\pi^2}{4}$

Here I attempt to calculate this integral ( only when $\alpha = 1$ ) without using $\zeta(s)$ or even this fact $\displaystyle \frac{1}{1^2} + \frac{1}{2^2} + .... = \frac{\pi^2}{6}$

First make the substitution $\displaystyle \sinh(x) = t$ or $\displaystyle x = \ln(t + \sqrt{t^2+1} )$ , then $\displaystyle dx = \frac{dt}{\sqrt{t^2+1} }$ The integral becomes :

$\displaystyle \int_0^{\infty} \frac{\ln(t+\sqrt{t^2+1} ) }{t\sqrt{t^2+1} } ~dt$

$\displaystyle = \int_0^{\infty} \ln{\left[\frac{\sqrt{t^2+1}+t}{\sqrt{t^2+1}-t}\right]} ~ \frac{dt}{2t\sqrt{t^2+1} }$

$\displaystyle = \int_0^{\infty} \int_0^1 \frac{dy}{t^2 + 1 - t^2y^2 } ~dt$

$\displaystyle = \int_0^1 \int_0^{\infty} \frac{dt}{t^2(1-y^2)+1}~dy$

$\displaystyle = \frac{\pi}{2} \int_0^1 \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle = \frac{\pi^2}{4}$

These three integrals come from American Mathematical Monthly (AMM E3140 )

Let $K = \frac{\ln^2(1+\sqrt{2})}{2}$
Show that :

$\displaystyle \int_1^{\infty} \frac{\ln(x+\sqrt{x^2 - 1} )}{x(x^2 +1 )}~dx = K$

$\displaystyle \int_1^{\infty}\frac{\tan^{-1}(x)}{x\sqrt{x^2 - 1 } }~dx = \frac{\pi^2}{8} + K$

$\displaystyle \int_0^1 \frac{\tan^{-1}(x)}{\sqrt{1-x^2} }~dx = \frac{\pi^2}{8} - K$
• Nov 10th 2010, 02:42 AM
tonio
Quote:

Originally Posted by melese
$\displaystyle\sum_{n=1}^\infty \frac{1}{(4n-1)(4n+1)}=\displaystyle\sum_{n=1}^\infty \frac{1}{4}(\frac{1}{4n-1}-\frac{1}{4n+1})=\displaystyle\frac{1}{4}\sum_{n=1} ^\infty (\frac{1}{4n-1}-\frac{1}{4n+1})=\displaystyle\frac{1}{4}\sum_{n=1} ^\infty (\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\cdots)$.

Using the Gregory–Leibniz series $\displaystyle\sum_{n=1}^\infty (\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\cdots)=\pi/4$ we find that the sum of the original sum is: $\frac{1}{4}(1-\frac{\pi}{4})$

After the first equality it must be $\frac{1}{2}$ , not $\frac{1}{4}$ , and the Gregory=Leibniz series is $1-\frac{1}{3}+\frac{1}{5}-\ldots=\frac{\pi}{4}$ ,

so the sum must in fact be, I think, $\frac{1}{2}\left(1-\frac{\pi}{4}\right)$

Tonio
• Nov 10th 2010, 05:31 PM
Bruno J.
Quote:

Originally Posted by simplependulum
I think most people will do this as the first step :

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

$\displaystyle = - \int_0^{1/2} \frac{ \ln(1-x) }{x}~dx$

Then , from $\displaystyle \frac{1}{x} + \frac{1}{1-x} = \frac{1}{x(1-x)}$

we have

$\displaystyle S = \int_0^{1/2} \frac{\ln(1-x)}{1-x}~dx - \int_0^{1/2} \frac{\ln(1-x)}{x(1-x)}~dx$

$\displaystyle = \left[-\frac{\ln^2(1-x)}{2} \right]_0^{1/2} - \int_0^{1/2} \frac{\ln(1-x)}{x(1-x)}~dx$

For the second integral , sub $\displaystyle x = \frac{t}{t+1}$ we find that

$\displaystyle \int_0^{1/2} \frac{\ln(1-x)}{x(1-x)}~dx = \int_0^1 \frac{-\ln(t+1)}{\frac{t}{(t+1)^2} }\cdot \frac{dt}{(t+1)^2}$

$\displaystyle = - \int_0^1 \frac{\ln(t+1)}{t}~dt$

Recall another famous integral expression for $\displaystyle \frac{\pi^2}{12} = \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - ... = \int_0^1 \frac{\ln(t+1)}{t}~dt$

Therefore , $\displaystyle S = \frac{\pi^2}{12} - \frac{\ln^2{2}}{2}$

All of the other solutions in this thread, while ingenious, are relatively straightforward. However you, simplependulum, always come up with an elementary and magical substitution which does the trick! Where do you find these ideas? First, partial fractions backwards, and then the innocent-looking but not obvious $x=t/(t+1)$... Please share some of your techniques for finding good substitutions (other than "practice", obviously (Rofl))!
• Nov 10th 2010, 06:17 PM
Drexel28
Quote:

Originally Posted by Bruno J.
All of the other solutions in this thread, while ingenious, are relatively straightforward. However you, simplependulum, always come up with an elementary and magical substitution which does the trick! Where do you find these ideas? First, partial fractions backwards, and then the innocent-looking but not obvious $x=t/(t+1)$... Please share some of your techniques for finding good substitutions (other than "practice", obviously (Rofl))!

If you're looking for tricky substitutions, you can do the integral I did (in the actual form you presented it) by making a couple non-obvious substitutions and recalling that $\sinh^{-1}(\frac{1}{x})=\cosh^{-1}(x)$. I'll let someone else try it out.
• Nov 10th 2010, 07:35 PM
Bruno J.
Quote:

Originally Posted by Drexel28
If you're looking for tricky substitutions, you can do the integral I did (in the actual form you presented it) by making a couple non-obvious substitutions and recalling that $\sinh^{-1}(\frac{1}{x})=\cosh^{-1}(x)$. I'll let someone else try it out.

Are you sure about that identity?
• Nov 10th 2010, 07:36 PM
Bruno J.
Quote:

Originally Posted by simplependulum
These three integrals come from American Mathematical Monthly (AMM E3140 )

Let $K = \frac{\ln^2(1+\sqrt{2})}{2}$
Show that :

$\displaystyle \int_1^{\infty} \frac{\ln(x+\sqrt{x^2 - 1} )}{x(x^2 +1 )}~dx = K$

$\displaystyle \int_1^{\infty}\frac{\tan^{-1}(x)}{x\sqrt{x^2 - 1 } }~dx = \frac{\pi^2}{8} + K$

$\displaystyle \int_0^1 \frac{\tan^{-1}(x)}{\sqrt{1-x^2} }~dx = \frac{\pi^2}{8} - K$

Have the solutions been given as well? I think I can almost solve the third one. It's pretty difficult but it's very nice!
• Nov 10th 2010, 10:08 PM
simplependulum
Quote:

Originally Posted by Bruno J.
All of the other solutions in this thread, while ingenious, are relatively straightforward. However you, simplependulum, always come up with an elementary and magical substitution which does the trick! Where do you find these ideas? First, partial fractions backwards, and then the innocent-looking but not obvious $x=t/(t+1)$... Please share some of your techniques for finding good substitutions (other than "practice", obviously (Rofl))!

For the first step to evaluate the integral , we can actually use the substitution followed by partial fractions , it becomes more obvious if we do this :

After the substitution , we will have

$\int_0^1 \frac{\ln(1+t)}{t(t+1)} ~dt$ , well , which is more obvious for us to think of $\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$ .

Quote:

Originally Posted by Bruno J.
Have the solutions been given as well? I think I can almost solve the third one. It's pretty difficult but it's very nice!

Talking about these three integrals (actually two , the ideas of the last two integrals are almost the same ) , I remember I'd spent a few weeks to finish when i was a fool who didn't realize the power of integration of parameter . Now , I know some basics of the techniques and these integrals look much easier to me BUT i have to say they are still difficult if we compare with many other integrals !!

I've got the solutions , later i will include them in this thread (Happy) .
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