Residue theorem VII

Nice integral involving ${}_1F_{1}$

Today we show the proof of this mysterious integral posted by @infseriesbot. $$\int_{0}^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx =\frac{\pi \Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)} {}_{1}F_{1}\left[{a-b\atop 1+a+b};z\right]$$ This integral resembles another one we previously solved using the Cauchy integral formula and it turned out that this also can be solved using contour integration.

Proof

First, note that the function $$f(x) = e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}$$ Is an even function...

Hence we can write: $$I = \int_{0}^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx = \frac{1}{2} \int_{-\pi }^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx$$ Expand the integrand using the Euler's formula: $$\begin{align*} I = \frac{1}{2}\int_{-\pi }^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx =& \frac{1}{2}\int_{-\pi }^{\pi}e^{-z\cos x}\left(\frac{e^{iax-iz\sin x}+e^{-iax+iz\sin x}}{2}\right)\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}dx\\ =& \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{ia-z(\cos x -i\sin x)} + e^{-iax +z(-\cos x +iz\sin x)}\right)\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}dx\\ =& \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-z(\cos x -i\sin x)} + e^{-iax}e^{+z(-\cos x +iz\sin x)}\right)\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}dx\\ =& \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-ze^{ix}} + e^{-iax}e^{+ze^{-ix}}\right)\underbrace{\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}}_{A}dx \end{align*}$$ Expand $A$ with the binomial theorem: $$A = \left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b} = \sum_{j=0}^{\infty} \binom{2b}{j} e^{ix(b-j)}$$ Hence $$I = \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-ze^{ix}} + e^{-iax}e^{+ze^{-ix}}\right)\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}dx = \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-ze^{ix}} + e^{-iax}e^{+ze^{-ix}}\right)\sum_{j=0}^{\infty} \binom{2b}{j} e^{ix(b-j)}dx$$ Now, do the following substitution: $$e^{ix} = w$$ $$dx = \frac{dw}{wi}$$ Our integral is transformed in a contour integral round the unit circle: $$\begin{align*} I = \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-ze^{ix}} + e^{-iax}e^{+ze^{-ix}}\right)\sum_{j=0}^{\infty} \binom{2b}{j} e^{ix(b-j)}dx =& \frac{1}{4}\oint_{|w|=1} \frac{\left( w^a e^{-zw} + w^{-a}e^{+zw^{-1}}\right)\sum_{j=0}^{\infty} \binom{2b}{j} w^{(b-j)}}{wi}dw\\ =&\frac{1}{4i}\oint_{|w|=1} \left( w^{a-1} e^{-zw} + w^{-a-1}e^{zw^{-1}}\right)\sum_{j=0}^{\infty} \binom{2b}{j} w^{(b-j)}dw\\ =& \frac{1}{4i}\underbrace{\oint_{|w|=1} \sum_{j=0}^{\infty} \binom{2b}{j} w^{(a+b-j-1)}e^{-zw}dw}_{J} + \frac{1}{4i}\underbrace{\oint_{|w|=1} \sum_{j=0}^{\infty} \binom{2b}{j} w^{(b-a-j-1)}e^{zw^{-1}}dw}_{K} \tag{1} \end{align*}$$ Expand $e^{-zw}, e^{-zw^{-1}}$ in $J$ and $K$ and apply the residue theorem: $$J = \oint_{|w|=1} \sum_{j=0}^{\infty} \binom{2b}{j} w^{(a+b-j-1)}e^{-zw}dw = \oint_{|w|=1} \underbrace{ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \binom{2b}{j} \frac{w^{(a+b-j+k-1)}(-z)^k}{k!}}_{g(w)}dw = 2\pi i \operatorname{Res}(g,0)$$ $$K = \oint_{|w|=1} \sum_{j=0}^{\infty} \binom{2b}{j} w^{(b-a-j-1)}e^{zw^{-1}}dw = \oint_{|w|=1} \underbrace{\sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \binom{2b}{j} \frac{w^{(b-a-j-k-1)}(-z)^n}{k!}}_{h(w)}dw = 2\pi i\operatorname{Res}(h,0)$$ The functions $g(w),h(w)$ have a singularity at $w=0$.

To find the residue we need to know the coefficient of $w^{-1}$

For $J$: $$w^{-1} = w^{a+b-j+k-1 } \Longrightarrow j = a+b+k$$ For $K$: $$w^{-1} = w^{b-a-j-k-1} \Longrightarrow j = b-a-k$$ Therefore $$J = 2\pi i\operatorname{Res}(g,0) = 2\pi i\sum_{k=0}^{\infty} \binom{2b}{a+b+k} \frac{(-z)^k}{k!}$$ $$K = 2\pi i\operatorname{Res}(h,0) = 2\pi i \sum_{k=0}^{\infty} \binom{2b}{b-a-k} \frac{(-z)^k}{k!}$$ Hence, from (1): $$I = \frac{1}{4i}J + \frac{1}{4i}K = \frac{\pi}{2}\sum_{k=0}^{\infty} \binom{2b}{a+b+k} \frac{(-z)^k}{k!} + \frac{\pi}{2} \sum_{k=0}^{\infty} \binom{2b}{b-a-k} \frac{(-z)^k}{k!}$$ Using the property $\displaystyle \binom{n}{n-m} = \binom{n}{m}$: $$\binom{2b}{b-a-k} = \binom{2b}{2b-a-b-k} = \binom{2b}{a+b+k}$$ Therefore $$I = \pi\sum_{k=0}^{\infty}\binom{2b}{a+b+k} \frac{(-z)^k}{k!} \tag{2}$$ From the formula that relates the binomial coefficient to the Pochhammer symbol $\displaystyle \binom{v}{m} = \frac{(v-m+1)_{m}}{m!}$ : $$\binom{2b}{a+b+k} = \frac{(b-a-k+1)_{a+b+k}}{(a+b+k)!} = \frac{(b-a-k+1)_{a+b+k}}{\Gamma(a+b+k+1)} = \frac{(b-a-k+1)_{a+b+k}}{\Gamma(a+b+1)(a+b+1)_{k}}$$ Now, from the rule $\displaystyle (x)_{n+m} = (x)_{n}(x+n)_{m}$ we have $$\binom{2b}{a+b+k} = \frac{(b-a-k+1)_{a+b+k}}{\Gamma(a+b+k+1)(a+b+1)_{k}} = \frac{(b-a-k+1)_{k}(b-a+1)_{a+b}}{\Gamma(a+b+k+1)(a+b+1)_{k}}$$ From definition $$(b-a+1)_{a+b} = \frac{\Gamma(2b+1)}{\Gamma(b-a+1)}$$ Therefore $$\binom{2b}{a+b+k} = \frac{(b-a-k+1)_{k}\Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)(a+b+1)_{k}}$$ Hence, from (2): $$\begin{align*} I = \int_{0}^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx = & \pi \sum_{j=0}^{\infty}\binom{2b}{a+b+k} \frac{(-z)^k}{k!}\\ =& \pi\sum_{j=0}^{\infty} \frac{(b-a-k+1)_{k}\Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)(a+b+1)_{k}} \frac{(-z)^{k}}{k!}\\ = & \frac{\pi\Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)}\sum_{k=0}^{\infty} \frac{(b-a-k+1)_{k}}{(a+b+1)_{k}} \frac{(-1)^{k}z^k}{k!}\\ =& \frac{\pi\Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)}\sum_{k=0}^{\infty} \frac{(a-b)_{k}}{(a+b+1)_{k}} \frac{z^k}{k!} \quad ((b-a-k+1)_{k}(-1)^k = (a-b)_{k}) \end{align*}$$ The last series satisfy the definition of ${}_{1}F_{1}$ also called Confluent hypergeometric function: $$\sum_{k=0}^{\infty} \frac{(a-b)_{k}}{(a+b+1)_{k}} \frac{z^k}{k!} = {}_{1}F_{1}\left[{a-b\atop 1+a+b};z\right]$$ Hence, we can conclude $$\boxed{ \int_{0}^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx =\frac{\pi \Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)} {}_{1}F_{1}\left[{a-b\atop 1+a+b};z\right]}$$