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Can \begin{equation} \sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left( 2k+1\right) ^{2}+a^{2}}, \end{equation} be summed explicitly, where $a$ is a constant real number? If $a=0,$ this sum becomes \begin{equation} \sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1}=\frac{\pi }{4}. \end{equation} What about for $a\neq 0$. I tried this method \begin{eqnarray*} \sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left( 2k+1\right) ^{2}+a^{2}} &=&\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}% \left[ \frac{1}{2k+1+ja}+\frac{1}{2k+1-ja}\right] \\ &=&\frac{1}{2}\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1+ja}+\frac{1}{2}% \sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1-ja} \\ &=&\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}\int_{0}^{1}x^{2k+ja}dx+% \frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}\int_{0}^{1}x^{2k-ja}dx \\ &=&\frac{1}{2}\int_{0}^{1}\left[ \sum_{k\geq 0}\left( -1\right) ^{k}x^{2k+ja}% \right] dx+\frac{1}{2}\int_{0}^{1}\left[ \sum_{k\geq 0}\left( -1\right) ^{k}x^{2k-ja}\right] dx \\ &=&\frac{1}{2}\int_{0}^{1}\frac{x^{ja}}{1+x^{2}}dx+\frac{1}{2}\int_{0}^{1}% \frac{x^{-ja}}{1+x^{2}}dx \\ &=&\frac{1}{2}\int_{0}^{1}\frac{x^{ja}+x^{-ja}}{1+x^{2}}dx \end{eqnarray*} but I was stucked at the last equations. Can any one give me some hint or tell me that the analytic expression doesn't exist. Thanks very much!

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Note for $x>0$: $x^{ia}+x^{-ia}=e^{ia\log(x)}+e^{-ia\log(x)}=2\cos(a\log(x))$. –  Your Ad Here Oct 31 '12 at 20:57
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3 Answers

up vote 5 down vote accepted

The answer is : $$S(a)=\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left( 2k+1\right) ^{2}+a^{2}}=\frac {\pi}{4\,\cosh(\pi a/2)}$$

Let's start nearly as you did : \begin{eqnarray*} S(a)=\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left( 2k+1\right) ^{2}+a^{2}} &=&\frac 12\sum_{k\geq 0}\left( -1\right) ^{k}% \left[ \frac 1{2k+1+ia}-\frac 1{-2k-1+ia}\right] \\ &=&\frac 12% \sum_{k=-\infty}^\infty \frac{\left( -1\right) ^{k}}{2k+1+ia} \\ &=&\frac 14% \sum_{k=-\infty}^\infty \frac{\left( -1\right) ^{k}}{k+(1+ia)/2} \\ \end{eqnarray*}

But the last series was evaluated earlier here using different methods with the result (in our case) : $$S(a)=\frac {\pi}{4\sin\left(\pi (1+ia)/2\right)}=\frac {\pi}{4\cos\left(\pi a i/2\right)}=\frac {\pi}{4\cosh\left(\pi a/2\right)}$$


Since your method is fine let's try to finish it your way :

$$ \begin{align} S(a)&=\frac 12\int_{0}^{1}\frac{x^{ia}+x^{-ia}}{1+x^{2}}dx\\ &=\int_{0}^{1}\frac{\cos(a\ln(x))}{1+x^{2}}dx\quad\text{using the hint in the comment}\\ &=\int_{-\infty}^0\frac{\cos(a\,t)\,e^t}{1+e^{2t}}dt\quad\text{if}\ \ x:=e^t\\ &=\frac 12\int_{-\infty}^0\frac{\cos(a\,t)}{\cosh(t)}dt\\ &=\frac 12\int_0^\infty \frac{\cos(a\,t)}{\cosh(t)}dt\\ \end{align} $$ Let's evaluate $\displaystyle I(a)=\int_{-R}^R \frac{e^{i a\,z}}{\cosh(z)}dz\ $ using a rectangular contour $[-R,+R]\times [0,\pi i]\ $ (we will have $R\to\infty$ and suppose $|a|<1$)

$$ I(a)=\int_{-R}^R \frac{e^{i a\,z}}{\cosh(z)}dz+\int_R^{R+\pi i} \frac{e^{i a\,z}}{\cosh(z)}dz+\int_{R+\pi i}^{-R+\pi i}\frac{e^{i a\,z}}{\cosh(z)}dz+\int_{-R+\pi i}^{-R} \frac{e^{i a\,z}}{\cosh(z)}dz $$ putting the third integral in second position : $$ I(a)=\int_{-R}^R \frac{e^{i a\,t}}{\cosh(t)}dt+\int_{R}^{-R}\frac{e^{i a(t+\pi i)}}{\cosh(t+\pi i)}dt+\int_R^{R+\pi i} \frac{e^{i a\,z}}{\cosh(z)}dz+\int_{-R+\pi i}^{-R} \frac{e^{i a\,z}}{\cosh(z)}dz $$ I'll let you show that the two integrals at the right disappear as $R\to\infty$ (for $|a|<1$) and use $\cosh(t+\pi i)=-\cosh(t)$ to get : $$I(a)=\int_{-\infty}^\infty \frac{e^{i a\,t}}{\cosh(t)}dt+\int_{-\infty}^\infty\frac{e^{i a(t+\pi i)}}{\cosh(t)}dt=\left(1+e^{-a\pi}\right)\int_{-\infty}^\infty \frac{e^{i a\,t}}{\cosh(t)}dt$$

Since there is only one pole at $\ t=\frac {\pi}2 i\ $ we may use the residu there to get : $$\frac{2\pi i\, e^{i a\,\pi i/2}/i}{\left(1+e^{-\pi a}\right)}=\frac{2\pi\, e^{-\pi a/2}}{\left(1+e^{-\pi a}\right)}=\frac{\pi}{\cosh(\pi a/2)}=\int_{-\infty}^\infty \frac{e^{i a\,t}}{\cosh(t)}dt$$ and conclude that indeed :

$$\frac 12\int_0^\infty \frac{\cos(a\,t)}{\cosh(t)}dt=\frac{\pi}{4\cosh(\pi a/2)}$$

Perhaps not as simple as you hoped...

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Can you tell me how to derive the result? –  Xiangyu Meng Oct 31 '12 at 23:19
    
@XiangyuMeng: yes I am redacting a more complete answer. –  Raymond Manzoni Oct 31 '12 at 23:21
    
I checked both of your answers. It requires that |a|<1. Does the result still hold for |a|>1? –  Xiangyu Meng Nov 1 '12 at 22:35
    
@XiangyuMeng: for the last derivation I supposed $|a|<1$ for convenience. Supposing that $|\Im{a}|<1$ should be enough (note that $a=i$ would give $0$ in your series' denominator) since the real part contributes only a $e^{i\Re{a}}$ part in the two neglected integrals at the right. Anyway you may use analytic extension to extend the rather smooth result as wished. –  Raymond Manzoni Nov 1 '12 at 23:07
    
I assume that $a$ is a real number, thus the |ℑa|<1 would be satisfied. Thanks very much for your answer. –  Xiangyu Meng Nov 1 '12 at 23:14
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This sum can be evaluated by the same trick as presented here: math.stackexchange.com.

In order to capture the convergence behavior of the series group consecutive terms to obtain absolute convergence, writing $$ S(a) = \sum_{m\ge 0} \frac{4m+1}{(4m+1)^2+a^2} - \sum_{m\ge 0} \frac{4m+3}{(4m+3)^2+a^2} = \sum_{m\ge 0} \frac{(4m+1)((4m+3)^2+a^2)-(4m+3)((4m+1)^2+a^2)} {((4m+3)^2+a^2)((4m+1)^2+a^2)} = \sum_{m\ge 0} \frac{2(4m+1)(4m+3)-2a^2}{((4m+3)^2+a^2)((4m+1)^2+a^2)} $$

Instead of using $f_1(z)$ and $f_2(z)$ from the other post use $$f(z) = \frac{2(4z+1)(4z+3)-2a^2}{((4z+3)^2+a^2)((4z+1)^2+a^2)} \pi \cot(\pi z).$$

The key operation of this technique is to compute the integral of $f(z)$ along a circle of radius $R$ in the complex plane, where $R$ goes to infinity. We certainly have $|\pi\cot(\pi z)|<2\pi$ for $R$ large enough. The core term from the sum is $\theta(R^2/R^4)$ which is $\theta(1/R^2)$ so that the integrals are $\theta(1/R)$ and vanish in the limit. This means that the sum of the residues at the poles of $f(z)$ add up to zero.

It is easily verified that the poles at the integers produce the terms of the sum twice over. It follows that $$ 2 S(a) + \text{Res}_{z=\frac{3}{4}-\frac{1}{4}ia} f(z) + \text{Res}_{z=\frac{3}{4}+\frac{1}{4}ia} f(z) + \text{Res}_{z=\frac{1}{4}-\frac{1}{4}ia} f(z) + \text{Res}_{z=\frac{1}{4}+\frac{1}{4}ia} f(z) = 0$$

The residues are easily computed as the poles are all simple. This finally yields $$ S(a) = \frac{1}{4} \frac{\pi}{\cosh \left(\frac{\pi a}{2}\right)}.$$

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The series is summable and has the closed form formula

$$-\frac{1}{4}\,{\frac {\pi }{\sin \left( \frac{1}{2}\pi \, \left( 3 + ia \right ) \right) }} = \frac{1}{4}{\frac {\pi }{\cosh \left( \frac{\pi \,a}{2} \right) }} .$$

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Can you tell me how to derive the result? –  Xiangyu Meng Oct 31 '12 at 23:19
    
@did: I hope you can see it now! –  Mhenni Benghorbal Nov 2 '12 at 19:02
    
@did:Lift up your downvote. –  Mhenni Benghorbal Nov 2 '12 at 19:03
    
@did:did: I do not know what to say, but if you do not know something, do not talk about it –  Mhenni Benghorbal Nov 2 '12 at 20:40
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