Integral of the day XV

Monstrous integrals posted by @infseriesbot

Today we show the proof of this triplet of very weird integrals posted by @infseriesbot here, here and here:

Let $|z|\lt1$ , then \[\int_{0}^{\pi} \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}} dx = \pi \] \[ \int_{0}^{\pi} \cos\left(bx -a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx = \pi \] \[\int_{0}^{\frac{\pi}{2}} \cos\left(bx-a\arctan\left(\frac{z\sin 2x}{1+z\cos 2x}\right)\right)(1+2z\cos2x +z^2)^{-\frac{a}{2}}\cos^{c-1}x dx = \frac{\pi}{2^c}\frac{\Gamma(c)}{\Gamma\left(\tfrac{1-b+c}{2}\right)\Gamma\left(\tfrac{1+b+c}{2}\right)}{{}_{2}F_{1}}\left[{a,\tfrac{1+b-c}{2}\atop \tfrac{1+b+c}{2}};z\right] \] These results are consequence of the residue theorem and some properties of the hypergeometric functions. We previously solved similar integrals here and here using similar techniques.

Proof:

1. First integral

Let \[I = \int_{0}^{\pi} \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}} dx \] First, note that the function \[f(x) = \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\] is an even function. Then, \[I = \frac{1}{2}\int_{-\pi}^{\pi} \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}} dx \] Using the fact that $\Re (e^{i\varphi}) = \cos(\varphi)$: \[ I = \Re\left( \frac{1}{2}\int_{-\pi}^{\pi} e^{\left(ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)}(1+2z\cos x+z^2)^{-\frac{a}{2}} dx\right)\] Making the following substitution: \[ e^{ix} = w\] \[ \cos(x) = \frac{w+w^{-1}}{2}\] \[ \sin(x) = \frac{w-w^{-1}}{2i}\] \[ dx = \frac{dw}{wi} \] Using the logarithmic definition of $\arctan$ and after a lengthy but easy calculation we can obtain: \begin{align*}\arctan\left(\frac{z\sin x}{1+z\cos x}\right) =& \arctan\left(\frac{i(z-w^2z)}{w^2z+2w+z}\right)\\ =& -\frac{i}{2}\ln\left(\frac{w^2z+w}{w+z}\right) \tag{1} \end{align*} \[e^{\left(ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)} = e^{\frac{a}{2}\ln\left(\frac{w^2z+w}{w+z}\right)} = \left(\frac{w^2z+w}{w+z}\right)^{\frac{a}{2}}\] Also note that \[(1+2z\cos x + z^2) = \frac{(w+z)(wz+1)}{w} \tag{2}\] Then, our integral became a contour integral round the unit complex circle: \[ I = \Re \left(\frac{1}{2}\oint_{|w|=1} \left(\frac{w^2z+w}{w+z}\right)^{\frac{a}{2}}\frac{w^{\frac{a}{2}}}{{(w+z)^{\frac{a}{2}}}(wz+1)^{\frac{a}{2}}wi} dw\right) = \Re \left(\frac{1}{2i}\oint_{|w|=1} \frac{w^{a-1}}{(w+z)^a} dw\right)\] If $|w|=1>|z|$, we can expand the denominator with the binomial theorem: \[ \Re \left(\frac{1}{2i}\oint_{|w|=1} \frac{w^{a-1}}{(w+z)^a} dw \right)= \Re\left( \frac{1}{2i}\oint_{|w|=1} \sum_{j=0}^{\infty} \binom{-a}{j} w^{-j-1}z^{j} \right) dw\] The function: \[ g(w) = \sum_{j=0}^{\infty}\binom{-a}{j} w^{-j-1}z^{j} \] Has a pole at $w=0$, then the residue is the coefficient of $w^{-1}$: \[ -j-1 = -1 \Longrightarrow j=0 \] \[ \operatorname{Res}\left(g(w),0\right) = 1 \] By the residue theorem: \[ \Re\left( \frac{1}{2i}\oint_{|w|=1} \sum_{j=0}^{\infty} w^{-j-1}z^{j} dw\right) = \Re \left(\pi \operatorname{Res}\left(g(w),0\right)\right) = \Re (\pi) = \pi \] Therefore \[\boxed{ \int_{0}^{\pi} \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}} dx = \pi} \]

2. Second integral:

\[ J = \int_{0}^{\pi} \cos\left(bx -a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx \] The integrand is also even, then \[ J = \frac{1}{2}\int_{-\pi}^{\pi} \cos\left(bx -a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx \] and using the fact that $\Re (e^{i\varphi}) = \cos(\varphi)$ \[ J = \Re\left(\frac{1}{2}\int_{-\pi}^{\pi} e^{\left(ibx -ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)}(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx\right) \] We can expand $\displaystyle \left(2\cos \frac{x}{2}\right)^{2b}$ with the binomial theorem \[ \left(2\cos \frac{x}{2}\right)^{2b} = (e^{\frac{ix}{2}}+ e^{\frac{-ix}{2}})^{2b} = \sum_{j=0}^{\infty} \binom{2b}{j} e^{ix(b-j)}\] Hence \[ J = \Re\left(\sum_{j=0}^{\infty} \binom{2b}{j} \frac{1}{2}\int_{-\pi}^{\pi} e^{\left(-ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)}(1+2z\cos x+z^2)^{-\frac{a}{2}}e^{ix(2b-j)}dx\right)\] If we make the substitution: \[ e^{ix} = w\] \[ dx = \frac{dw}{wi} \] and using (1) and (2). We have: \[e^{\left(-ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)} = \left(\frac{w+z}{w^2z+w}\right)^{\frac{a}{2}}\] and \[(1+2z\cos x + z^2) = \frac{(w+z)(wz+1)}{w}\] Hence, \begin{align*} J =& \Re\left(\sum_{j=0}^{\infty} \binom{2b}{j} \frac{1}{2i}\oint_{|w|=1} \left(\frac{w+z}{w^2z+w}\right)^{\frac{a}{2}}\frac{w^{\frac{a}{2}}}{(w+z)^{\frac{a}{2}}(wz+1)^{\frac{a}{2}}w} w^{(2b-j)}dw\right)\\ =& \Re\left(\sum_{j=0}^{\infty} \binom{2b}{j} \frac{1}{2i}\oint_{|w|=1} \frac{w^{2b-j-1}}{(wz+1)^a}dw\right)\\ \end{align*} If $|z|\lt1$ \[ \frac{1}{(wz+1)^a} = \sum_{k=0}^{\infty} \binom{-a}{k}w^kz^k \] Hence, \begin{align*}J=& \Re\left(\sum_{j=0}^{\infty} \binom{2b}{j} \frac{1}{2i}\oint_{|w|=1} \sum_{k=0}^{\infty} \binom{-a}{k}w^{2b-j+k-1}z^k dw\right)\\ =& \Re\left(\sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \binom{2b}{j} \binom{-a}{k}\frac{z^k }{2i}\oint_{|w|=1} w^{2b-j+k-1}dw\right) \end{align*} By the residue theorem \begin{align*} J =& \Re\left(\sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \binom{2b}{j} \binom{-a}{k}\frac{z^k }{2i}\oint_{|w|=1} w^{2b-j+k-1}dw\right) \\ =&\Re\left(\pi \sum_{k=0}^{\infty} \binom{2b}{2b+k} \binom{-a}{k}z^k\right) \end{align*} Note that \[ \binom{2b}{2b+k} = 0 \Longleftrightarrow k>0 \wedge b>0 \] Therefore \[ J = \Re\left(\pi \sum_{k=0}^{\infty} \binom{2b}{2b+k} \binom{-a}{k}z^k\right) = \pi \binom{-a}{0}\binom{2b}{2b} = \pi \] Hence, we can conclude \[ \boxed{\int_{0}^{\pi} \cos\left(bx -a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx = \pi }\]

3. Third integral

Finally, we evaluate the last monstrous integral of this triplet using the results obtained in the previous integrals. Additionally, we need to make the assumption that $\displaystyle \tfrac{c-1+b}{2} \in \mathbb{N}$ \[ K = \int_{0}^{\frac{\pi}{2}} \cos\left(bx-a\arctan\left(\frac{z\sin 2x}{1+z\cos 2x}\right)\right)(1+2z\cos2x +z^2)^{-\frac{a}{2}}\cos^{c-1}x dx \] Do the substitution $\displaystyle t = 2x$. \[ K = \frac{1}{2} \int_{0}^{\pi} \cos\left(\frac{bt}{2}-a\arctan\left(\frac{z\sin t}{1+z\cos t}\right)\right)(1+2z\cos t +z^2)^{-\frac{a}{2}}\cos^{c-1}\left(\frac{t}{2}\right) dt\] Again, the integrand is an even function of $t$: \[ K = \frac{1}{4} \int_{-\pi}^{\pi} \cos\left(\frac{bt}{2}-a\arctan\left(\frac{z\sin t}{1+z\cos t}\right)\right)(1+2z\cos t +z^2)^{-\frac{a}{2}}\cos^{c-1}\left(\frac{t}{2}\right) dt\] Using the fact that $\displaystyle \Re(e^{i\varphi}) = \cos(\varphi)$ \[ K = \Re \left(\frac{1}{4} \int_{-\pi}^{\pi} e^{\left(i\frac{bt}{2}-ia\arctan\left(\frac{z\sin t}{1+z\cos t}\right)\right)}(1+2z\cos t +z^2)^{-\frac{a}{2}}\cos^{c-1}\left(\frac{t}{2}\right) dt\right)\] Expanding $\displaystyle \cos^{c-1}\left(\frac{t}{2}\right)$ with the binomial theorem: \[ K = \Re \left(\sum_{j=0}^{\infty} \binom{c-1}{j} \frac{1}{2^{c+1}} \int_{-\pi}^{\pi} e^{\left(-ia\arctan\left(\frac{z\sin t}{1+z\cos t}\right)\right)}(1+2z\cos t +z^2)^{-\frac{a}{2}}e^{it\left(\frac{b+c-1}{2}-j\right)} dt\right)\] Again, making the substitution \[ e^{it} = w\] \[ dt = \frac{dw}{wi} \] and using (1) and (2) we have \begin{align*} K = &\Re \left(\sum_{j=0}^{\infty} \binom{c-1}{j} \frac{1}{2^{c+1}i} \oint_{|w|=1}\left(\frac{w+z}{w^2z+w}\right)^{\frac{a}{2}}\frac{w^{-\frac{a}{2}}}{(w+z)^{\frac{a}{2}}(wz+1)^{\frac{a}{2}}w}w^{\left(\frac{b+c-1}{2}-j\right)} dt\right)\\ =& \Re \left(\sum_{j=0}^{\infty} \binom{c-1}{j} \frac{1}{2^{c+1}i} \oint_{|w|=1} \frac{w^{\left(\frac{b+c-1}{2}-j-1\right)}}{(wz+1)^a} dw\right) \end{align*} Assuming that $|z|\lt1$ and expanding $\displaystyle \frac{1}{(wz+1)^a}$ with the binomial theorem: \begin{align*} K=& \Re \left(\sum_{j=0}^{\infty} \binom{c-1}{j} \frac{1}{2^{c+1}i} \oint_{|w|=1} \sum_{k=0}^{\infty} \binom{-a}{k} w^{\left(\frac{b+c-1}{2}-j+k-1\right)}z^k dw\right)\\ =&\Re \left(\sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \binom{c-1}{j} \binom{-a}{k} \frac{z^k}{2^{c+1}i} \oint_{|w|=1} w^{\left(\frac{b+c-1}{2}-j+k-1\right)} dw\right) \end{align*} Applying the residue theorem: \[ K = \Re \left(\sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \binom{c-1}{j} \binom{-a}{k} \frac{z^k}{2^{c}}\pi \operatorname{Res}\left(w^{\left(\frac{b+c-1}{2}-j+k-1\right)},0\right)\right)\] The residue is the coefficient of $w^{-1}$ Then \[ w^{-1} = w^{\left(\frac{b+c-1}{2}-j+k-1\right)} \Longrightarrow j = k+\frac{b+c-1}{2}\] Since all the parameters are real or natural numbers we can dropout the $\Re$. Therefore, \[ K = \frac{\pi}{2^c}\sum_{j=0}^{\infty} \binom{c-1}{ k+\frac{b+c-1}{2}} \binom{-a}{k} z^{k}\]

Who is this mysterious function? Let's see

Recall the formula that relate the binomial coefficient to the Pochhammer polynomial or rising factorial: \[ \binom{v}{m} = \frac{(v-m+1)_{m}}{m!} \] Hence \begin{align*} K =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \binom{c-1}{ k+\frac{b+c-1}{2}} \binom{-a}{k} z^{k} \\ &= \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{\left(\tfrac{c-1-b}{2}-k+1\right)_{k+\tfrac{b+c-1}{2}}(-a-k+1)_{k}}{\left(k+\tfrac{b+c-1}{2}\right)!k!} z^{k} \end{align*} Now, using the rule for rising factorial \[ (x)_{n+m} = (x)_{n}(x+n)_{m} \] we have \[\left(\tfrac{c-1-b}{2}-k+1\right)_{k+\tfrac{b+c-1}{2}} = \left(\tfrac{c-1-b}{2}-k+1\right)_{k}\left(\tfrac{c-1-b}{2}+1\right)_{\tfrac{b+c+1}{2}} \tag{3}\] Now, using the reflection formula \[(-x)_{n} = (-1)^n(x-n+1)_{n} \tag{4} \] and the fact that the rising factorial can be expressed as the quotient of two gamma functions: \[ (x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)} \tag{5}\] we have \begin{align*} K= &\frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{\left(\tfrac{c-1-b}{2}-k+1\right)_{k+\tfrac{b+c-1}{2}}(-a-k+1)_{k}}{\left(k+\tfrac{b+c-1}{2}\right)!k!} z^{k}\\ =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{ \left(\tfrac{c-1-b}{2}-k+1\right)_{k}\left(\tfrac{c-1-b}{2}+1\right)_{\tfrac{b+c+1}{2}}(-a-k+1)_{k}}{\left(k+\tfrac{b+c-1}{2}\right)!k!} z^{k} \quad \textrm{ from } (3)\\ =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{(-1)^k\left(\tfrac{b+1-c}{2}\right)_{k}\Gamma(c)(-a-k+1)_{k}}{\Gamma\left(\tfrac{c+1-b}{2}\right)\left(k+\tfrac{b+c-1}{2}\right)!k!} z^{k} \quad \textrm{ from } (4),(5)\\ =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{\left(\tfrac{b+1-c}{2}\right)_{k}\Gamma(c)(a)_{k}}{\Gamma\left(\tfrac{c+1-b}{2}\right)\Gamma\left(k+\tfrac{b+c+1}{2}\right)k!} z^{k}\\ =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{\left(\tfrac{b+1-c}{2}\right)_{k}\Gamma(c)(a)_{k}}{\Gamma\left(\tfrac{c+1-b}{2}\right)\Gamma\left(\tfrac{b+c+1}{2}\right)\left(\tfrac{b+c+1}{2}\right)_{k}k!} z^{k}\\ =& \frac{\pi}{2^c}\frac{\Gamma(c)}{\Gamma\left(\tfrac{c+1-b}{2}\right)\Gamma\left(\tfrac{b+c+1}{2}\right)}\sum_{j=0}^{\infty} \frac{\left(\tfrac{b+1-c}{2}\right)_{k}(a)_{k}}{\left(\tfrac{b+c+1}{2}\right)_{k}} \frac{z^{k}}{k!}\\ =& \frac{\pi}{2^c}\frac{\Gamma(c)}{\Gamma\left(\tfrac{1-b+c}{2}\right)\Gamma\left(\tfrac{1+b+c}{2}\right)}{{}_{2}F_{1}}\left[{a,\tfrac{1+b-c}{2}\atop \tfrac{1+b+c}{2}};z\right]\\ \end{align*} Therefore \[ \boxed{\int_{0}^{\frac{\pi}{2}} \cos\left(bx-a\arctan\left(\frac{z\sin 2x}{1+z\cos 2x}\right)\right)(1+2z\cos2x +z^2)^{-\frac{a}{2}}\cos^{c-1}x dx = \frac{\pi}{2^c}\frac{\Gamma(c)}{\Gamma\left(\tfrac{1-b+c}{2}\right)\Gamma\left(\tfrac{1+b+c}{2}\right)}{{}_{2}F_{1}}\left[{a,\tfrac{1+b-c}{2}\atop \tfrac{1+b+c}{2}};z\right]}\]