Integral of the day XVIII

Integral involving a discrete Laplace transform

Today we evaluate the following integral posted by @integralsbot \[ \int_{0}^{\frac{\pi}{2}} \frac{\cos (\alpha x) \cos^{\alpha} (x)}{\cos (2\beta) - \cos (2x)} dx = \frac{\pi}{4e^{\alpha\beta}\sinh(\beta) \cosh^{1-\alpha}(\beta)} \quad \alpha >-1, \beta>0 \] The trick to evaluate this integral is the use of a discrete Laplace transform.

Proof.

This is the discrete Laplace transform that we need (Proof in the appendix): \[ 1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx = \frac{\sinh t}{\cosh t - \cos x} \quad { t>0 } \tag{1}\] Therefore \begin{align*} I = \int_{0}^{\frac{\pi}{2}} \frac{\cos (\alpha x) \cos^{\alpha} (x)}{\cos (2\beta) - \cos (2x)} dx = & \frac{1}{2}\int_{0}^{\pi} \frac{\cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right)}{\cos (2\beta) - \cos (w)} dw \quad (w \mapsto 2x) \\ =& \frac{1}{4}\int_{-\pi}^{\pi} \frac{\cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right)}{\cos (2\beta) - \cos( w)} dw \quad \textrm{ (the integrand is even)}\\ =& \frac{1}{4}\int_{-\pi}^{\pi} \frac{\sinh (2\beta)}{\cos (2\beta) - \cos( w)}\frac{\cos \left( \frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right)}{\sinh (2\beta )} dw \\ =& \frac{1}{4}\int_{-\pi}^{\pi}\left(1+2\sum_{k=1}^{\infty} e^{-k2\beta}\cos (kw) \right)\frac{\cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right)}{\sinh (2\beta) } dw \quad \textrm{ from (1) } \\ =& \frac{1}{4\sinh (2\beta)}\underbrace{\int_{-\pi}^{\pi}\cos \left(\alpha \frac{w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right) dw}_{J} + \frac{1}{2\sinh (2\beta)}\sum_{k=1}^{\infty} e^{-k 2\beta}\underbrace{\int_{-\pi}^{\pi} \cos(kw) \cos \left(\alpha \frac{w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right) dw}_{K} \tag{2} \\ \end{align*} Lets start with $J$. Using the fact that $ \Re(e^{ix}) = \cos(x)$: \[ J = \int_{-\pi}^{\pi}\cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right) dw = \Re\left(\int_{-\pi}^{\pi} e^{i\frac{\alpha w}{2}} \cos^{\alpha} \left(\frac{w}{2}\right) dw \right)\] If we make the substitution \[ e^{iw} = z\] \[ dw = \frac{dz}{zi} \] we transform the integral in a contour integral round the unit complex circle: \begin{align*} J = \Re\left(\int_{-\pi}^{\pi} e^{i\frac{\alpha w}{2}} \cos^{\alpha} \left(\frac{w}{2}\right) dw \right) =& \Re\left(\frac{1}{i} \oint_{|z|=1} \frac{z^{\frac{\alpha}{2}}(z+1)^{\alpha} }{2^{\alpha}z^{\frac{\alpha}{2}}z} dz\right)\\ =& \Re\left(\frac{1}{2^{\alpha}i} \oint_{|z|=1} \frac{(z+1)^{\alpha} }{z} dz\right) \end{align*} The function $ \displaystyle f(z) = \frac{(z+1)^{\alpha} }{z}$ has two singularities at $z=0$ and $z=-1$. However, since $\alpha >-1$, $z=-1$ is a removable singularity and its residue is zero. Therefore we can use the binomial theorem to find the residue at $z=0$: \[ (1+z)^{\alpha} = \sum_{j=0}^{\infty} \binom{\alpha}{j} z^{j} \quad |z|\lt 1 \] \[ f(z) = \frac{(z+1)^{\alpha} }{z} = \sum_{j=0}^{\infty} \binom{\alpha}{j} z^{j-1} \] The residue is the coefficient of $z^{-1}$ which means that $j=0$. By the Residue theorem: \[ \oint_{|z|=1} \frac{(z+1)^{\alpha} }{z} dz = 2\pi i \operatorname{Res}(f,0) = 2\pi i \binom{\alpha}{0} = 2\pi i \] Hence \[ J = \Re\left(\frac{1}{2^{\alpha}i} \oint_{|z|=1} \frac{(z+1)^{\alpha} }{z} dz\right) = \Re\left( \frac{\pi}{2^{\alpha-1}} \right) = \frac{\pi}{2^{\alpha-1}} \tag{3} \] Now for $K$: \begin{align*} K = \int_{-\pi}^{\pi} \cos(kw) \cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right) dw = \Re\left( \int_{-\pi}^{\pi} \cos(kw) e^{\frac{i\alpha w}{2}} \cos^{\alpha} \left(\frac{w}{2}\right) dw\right) \end{align*} Making the same substitution as before: \[ e^{iw} = z\] \[ dw = \frac{dz}{zi} \] \begin{align*} K = \Re\left( \int_{-\pi}^{\pi} \cos(kw) e^{\frac{i\alpha w}{2}} \cos^{\alpha} \left(\frac{w}{2}\right) dw\right) = \Re\left( \frac{1}{2^{\alpha+1}i}\oint_{|z|=1} \frac{(z^{2k}+1)(z+1)^{\alpha}}{z^{k+1}} dz\right) \end{align*} The function \[ g(z)= \frac{(z^{2k}+1)(z+1)^{\alpha}}{z^{k+1}} \] has two singularities at $z=-1$ and $z=0$ but again the singularity at $z=-1$ is a removable singularity and its residue is zero. To find the residue at $z=0$ we will use the binomial expansion of $(z+1)^{\alpha}$. \[ g(z)= \frac{(z^{2k}+1)}{z^{k+1}}\sum_{j=0}^{\infty} \binom{\alpha}{j} z^j = \sum_{j=0}^{\infty} \binom{\alpha}{j} z^{j+k-1} + \sum_{j=0}^{\infty} \binom{\alpha}{j} z^{j-k-1} \quad |z|\lt 1 \] Therefore the residue is the coefficient of $z^{-1}$. Note that the residue in the first sum is zero while the second sum attains the residue at $j=k$ \[ \oint_{|z|=1} \frac{(z^{2k}+1)(z+1)^{\alpha}}{z^{k+1}} dz = 2\pi i \operatorname{Res}(g,0) = 2\pi i \binom{\alpha}{k}\] Therefore \[ K = \Re\left( \frac{1}{4i}\oint_{|z|=1} \frac{(z^{2k}+1)(z+1)^{\alpha}}{z^{k+1}} dz\right) = \Re\left( \frac{\pi}{2^{\alpha}}\binom{\alpha}{k} \right) = \frac{\pi}{2^{\alpha}}\binom{\alpha}{k} \tag{4} \] Hence \begin{align*} I = \int_{0}^{\frac{\pi}{2}} \frac{\cos \alpha x \cos^{\alpha} x}{\cos 2\beta - \cos 2x} dx =& \frac{\pi}{2^{\alpha+1}\sinh (2\beta) } + \frac{\pi}{2^{\alpha+1}\sinh (2\beta)}\sum_{k=1}^{\infty}\binom{\alpha}{k} e^{-k 2\beta} \quad \textrm{ from (2),(3),(4)} \\ =& \frac{\pi}{2^{\alpha+1}\sinh (2\beta)}\sum_{k=0}^{\infty}\binom{\alpha}{k} e^{-k 2\beta} \quad \textrm{ (re-indexing) } \\ =& \frac{\pi}{2^{\alpha+1}\sinh (2\beta)}(1+e^{-2\beta})^{\alpha} \quad \textrm{ (binomial theorem) } \\ =& \frac{\pi}{2 e^{\alpha\beta}\sinh (2\beta)}\left(\frac{e^{\beta}+e^{-\beta}}{2}\right)^{\alpha} \\ =& \frac{\pi \cos^{\alpha}(\beta)}{2e^{\alpha\beta}\sinh (2\beta)} \\ =& \frac{\pi}{4 e^{\alpha\beta}\sinh(\beta)\cosh^{1-\alpha}(\beta)} \quad \left(\sinh(2\beta) = 2\sinh(\beta)\cosh(\beta)\right)\\ \end{align*} Therefore, we can conclude \[ \boxed{ \int_{0}^{\frac{\pi}{2}} \frac{\cos (\alpha x) \cos^{\alpha} (x)}{\cos (2\beta) - \cos (2x)} dx = \frac{\pi}{4e^{\alpha\beta}\sinh(\beta) \cosh^{1-\alpha}(\beta)} \quad \alpha >-1, \beta>0 } \] Appendix.

Proposition. \[ 1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx = \frac{\sinh t}{\cosh t - \cos x} \quad { t>0 }\] Proof: \begin{align*} 1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{-kt}e^{ikx} \right)\\ =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{k(ix-t)} \right)\\ =& 1+2\Re\left(\frac{1}{e^{t-ix}-1} \right)\\ =& 1+2\Re\left(\frac{e^{-t}}{e^{-ix}-e^{-t}} \right)\\ =& 1+2e^{-t}\Re\left(\frac{1}{\cos x-i\sin x- e^{-t}}\right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{(\cos x-e^{-t})^2 + \sin^2 x} \right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{e^{-2t}-2e^{t}\cos x + 1} \right)\\ =& 1+\Re\left(\frac{\cos x -e^{-t}+ i\sin x}{\cosh t-\cos x} \right)\\ =& \frac{\cosh t- \cos x}{\cosh t- \cos x}+\frac{\cos x-e^{-t}}{(\cosh t- \cos x )} \\ =& \frac{\sinh t}{\cosh t - \cos x} \end{align*}