# Thread: Real integral via the Residue Theorem

1. ## Real integral via the Residue Theorem

This is a challenge problem I already know how to solve.

Show via the Residue Theorem and without using Bessel functions or any of that other stuff in the other thread:

$\int_{-1}^{1}\frac{\cosh(2z)}{\sqrt{1-z^2}}dz=\pm\pi \sum_{k=0}^{\infty} \frac{1}{(k!)^2}$

with the $\pm$ due to the particular sheet of the underlying Riemann surface the integral is being integrated over.

2. But i find that this method is okay to me .

$\cosh(2z) = \sum_{k\geq 0} \frac{(2z)^{2k}}{(2k)!}$

Sub $z = \sqrt{t}$

$I = \sum_{k\geq 0} \frac{4^k}{(2k)!} \int_0^1 z^{k-1/2}(1-z)^{-1/2}~dz$

$= \sum_{k\geq 0} \frac{4^k}{(2k)!} \frac{ \Gamma(k+1/2) \Gamma(1/2)}{ \Gamma(k+1) }$

$= \sum_{k\geq 0} \frac{4^k}{(2k)!} \frac{ (2k)! \pi }{ 4^k (k!) (k!) }$

$= \pi \sum_{k\geq 0} \frac{1}{(k!)^2}$

3. Originally Posted by simplependulum
But i find that this method is okay to me .

$\cosh(2z) = \sum_{k\geq 0} \frac{(2z)^{2k}}{(2k)!}$

Sub $z = \sqrt{t}$

$I = \sum_{k\geq 0} \frac{4^k}{(2k)!} \int_0^1 z^{k-1/2}(1-z)^{-1/2}~dz$

$= \sum_{k\geq 0} \frac{4^k}{(2k)!} \frac{ \Gamma(k+1/2) \Gamma(1/2)}{ \Gamma(k+1) }$

$= \sum_{k\geq 0} \frac{4^k}{(2k)!} \frac{ (2k)! \pi }{ 4^k (k!) (k!) }$

$= \pi \sum_{k\geq 0} \frac{1}{(k!)^2}$
Nice. Didn't know how to do that. I'll post the residue computation in a few days if no one replies.

4. Hi. I cannot adequately justify the step below marked with an asterisk and thus feel I should not have posted this problem here. I apologize to the readers of this thread for that. I ask the moderators to please either leave it open here or move it to the general forum in hopes someone more familiar with the subject can review my work and suggest a means of improving it.

This problem concerns a particular case of the more general expression:

$\mathop\oint\limits_{|z|>H} E(z) P_n(z)^{\pm 1/n}dz,\quad n\in\mathbb{Z^+}$

for $E(z)$ entire and $P_n(z)$ an n'th degree polynomial where the contour is outside the convex hull of the zeros of the polynomial. In this case, the integrand has a set of n analytic branches, for which the integral over each can be easily computed using the Residue Theorem if care is taken to define each branch.

For this particular problem, let $f(z)=E(z)h(z)=\cosh(2z)\frac{1}{\sqrt{1-z^2}}$

Now what causes the difficulty with such problems is the multi-valued component $h(z)$ so part of the analysis and the plots below focus only on $h(z)$ since $\cosh(2z)$ is entire and will not change qualitatively, the analysis. The real component of $h(z)$ is shown in the first picture. Now consider the two contours in the second picture over a particular branch of $h(z)$ in the third picture which I'll call $A_2(h)$ so chosen so that along the red contour, $A_2(h)=h(x)>0$. We can construct this branch in the usual way and arrive at the expression:
$A_2(h)=\frac{1}{(r_1 r_2)^{1/2}e^{i/2(\theta_1+\theta_2+2\pi)}}$

Now by the Residue Theorem, traversing the inner contour in the negative sense is the negative of traversing the yellow contour in the positive sense or:
$\mathop\oint\limits_{\text{-inner}}A_2(h)dz=-\mathop\oint\limits_{\text{+yellow}}A_2(h)dz$

However, near the branch-cut, $A_2(h)\equiv \pm h(x)$ and this branch is chosen so that the red leg of the inner contour corresponds to the positive component of $\sqrt{1-z^2}$.

Then we have:

$\int_{-1}^1 f(x)dx+\int_{1}^{-1} (-f(x))dx+\mathop\int\limits_{\Gamma_1+\Gamma_2} f(z)dz=-\mathop\oint\limits_{\text{+yellow}} g(z)A_2(f)dz$

where $\Gamma_1+\Gamma_2$ are the brown and red contours around the poles. As the radii over $\Gamma_1$ and $\Gamma_2$ go to zero, the integrals over these contours approach $\pi i$ and $-\pi i$ so cancel. Since $A_2(h)$ is analytic outsize the branch cut, I can then use the residue at infinity to compute the integral over $g(z)A_2(h)$ and write:

$2\int_{-1}^1 f(x)dx=-\mathop\oint\limits_{|z|>1} g(z)A_2(h)dz=-2\pi i\mathop\text{Res}\limits_{z=0}\left\{\frac{1}{z^2 }(\cosh(2/z)A_2(h(1/z))\right\}$

Now the issue becomes how is that residue computed?

Note first $A_2(h)=\frac{1}{\sqrt{1-z^2}}$ as long as the root object is interpreted appropriately so if I can justify the following algebraic manipulations and pull out a $\sqrt{-1}=-i$ from the denominator, obtain:

\begin{aligned}
2\int_{-1}^1 f(x)dx=-\mathop\oint\limits_{|z|>1} g(z)A_2(h)dz&=-2\pi i\mathop\text{Res}\limits_{z=0}\left\{\frac{1}{z^2 }\frac{\cosh(2/z)}{\sqrt{1-\frac{1}{z^2}}}\right\}\\
&=2\pi\mathop\text{Res}\limits_{z=0}\left\{\frac{1 }{z}\frac{\cosh(2/z)}{\sqrt{(1-z^2)}}\right\}
\end{aligned}

That expression has an essential singularity at zero so I can't use any of the standard techniques of evaluating the residue if it were a pole, however, I can expand:

$\frac{1}{z}\cosh(2/z)=\sum_{n=0}^{\infty} \frac{2^{2n}}{z^{2n+1}(2n)!}$

$(1-z^2)^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2+n}{-1/2} z^{2n}$

so that:

$\mathop\text{Res}\limits_{z=0}\left\{\frac{\cosh(2/z}{z\sqrt{1-z^2}}\right\}=\mathop\text{Res}\limits_{z=0}\left\ {\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{2^{2k}}{z^{2k+1}(2k)!}\binom{-1/2+n-k}{-1/2}z^{2(n-k)}\right\}.$

Now the residue is the sum of the coefficients of $1/z$ or those when $2n-2k-(2k+1)=-1$ or $n=2k$ and therefore:


\begin{aligned}
\mathop\text{Res}\limits_{z=0}\left\{\frac{\cosh(2/z}{z\sqrt{1-z^2}}\right\}&=\sum_{k=0}^{\infty} \frac{2^{2k}}{(2k)!}\binom{-1/2+k}{-1/2} \\
&=\sum_{k=0}^{\infty}\frac{2^{2k}\Gamma(1/2+k)}{\Gamma(1/2)\Gamma(k+1)(2k)!}\\
&=\sum_{k=0}^{\infty}\frac{1}{(k!)^2}
\end{aligned}

using the identities $\Gamma(1/2)=\sqrt{\pi}$ and $\Gamma(1/2+k)=\frac{\Gamma(2k)\sqrt{\pi}}{2^{2k-1}\Gamma(k)}$.

And thus:

$\int_{-1}^{1}\frac{\cosh(2x)}{\sqrt{1-x^2}}dx=\pi\sum_{k=0}^{\infty}\frac{1}{(k!)^2}$