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I have recently been studying the Residue Formula, and I am puzzled about how to solve this integral. I can only solve the case where the upper limit of integration is $\pi$. Can this integral be solved using complex analysis?(In this integration, a, p are both real numbers, and $a>0$)

ty h
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    How did you evaluate the integral when the upper limit is $\pi$?? – Mark Viola Oct 24 '24 at 18:21
  • This is Gradshteyn and Ryzhik,Corrected and Enlarged edition, 3.631.9. The result is in terms of the Beta function. Are you familiar with the Beta function? Also, the $[0,\pi]$ version is 3.631.17, which, though elementary, would be surprising to obtain by hand. – Eric Towers Oct 24 '24 at 18:37
  • Have you tried $e^{ipx}=(\cos(x)+i\sin(x))^p$ (for suitable $p$ and then by analytic continuation) and taking the real part and using the change of variable $u=\tan(x)$? – Diger Oct 24 '24 at 18:45
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    https://math.stackexchange.com/questions/253381/evaluate-cosine-power-reduction-formula-in-terms-of-gamma-function?noredirect=1 – Svyatoslav Oct 24 '24 at 19:51
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    Also contained in my answer here. – metamorphy Oct 25 '24 at 07:10

2 Answers2

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\left.\int_{0}^{\pi/2}\cos^{a}\pars{x} \cos\pars{px}\,\dd x \right\vert_{\substack{a, p\ \in\ \mathbb{R}\\[1mm] \verts{p}\ >\ a\ >\ -1}}} \\[5mm] = & \ \Re\int_{0}^{\pi/2}\cos^{a}\pars{x}\expo{\ic\verts{p}x} \,\dd x \\[5mm] = & \ \left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2} \pars{z + 1/z \over 2}^{a}z^{\verts{p}} \,\,{\dd z \over \ic z}\right\vert_{z\ \equiv\ \exp\pars{\ic x}} \\[5mm] = & \ \left.{1 \over 2^{a}}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2} \pars{1 + z^{2}}^{a}\,z^{\verts{p} - a - 1} \,\,\,\dd z\right\vert_{z\ \equiv\ \exp\pars{\ic x}} \\[5mm] = & \ -{1 \over 2^{a}}\,\Im\int_{1}^{0} \pars{1 - y^{2}}^{a}\, y^{\verts{p} - a - 1}\,\, \expo{\ic\pars{\verts{p} - a - 1}\pi/2} \,\,\,\,\ic\,\dd y \\[5mm] = & \ {1 \over 2^{a}}\sin\pars{\bracks{\verts{p} - a}{\pi \over 2}}\int_{0}^{1} \pars{1 - y^{2}}^{a}\, y^{\verts{p} - a - 1}\,\,\,\,\,\,\dd y \\[5mm] \sr{y^{2}\ \mapsto\ y}{=} & \ {\sin\pars{\bracks{\verts{p} - a}\pi/2} \over 2^{a + 1}} \int_{0}^{1}y^{\verts{p}/2 - a/2 - 1}\,\,\,\, \pars{1 - y}^{a}\,\,\dd y \\[5mm] = & \ \bbx{\color{#44f}{{\sin\pars{\bracks{\verts{p} - a}\pi/2} \over 2^{a + 1}}{\Gamma\pars{\verts{p}/2 - a/2}\Gamma\pars{a + 1} \over \Gamma\pars{\verts{p}/2 + a/2 + 1}}}} \\ & \end{align} Note that the condition $\ds{\verts{p} > a > -1}$ yields the last integral convergence. However, the $\lim_{a\ \to\ \verts{p}}$ of the final result becomes $\ds{\pi/2^{\verts{p} + 1}}$.

Felix Marin
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Here is a sketch. Let $I$ be the integral. It can be written as $$I=\frac12 \int_{-\infty}^\infty (1+iz)^{p/2-a/2-1}(1-iz)^{-p/2-a/2-1} \, dz \, .$$ In the upper half plane the integrand has a cut along $(i,i\infty)$. Furthermore, the integral along the arc $Re^{it}$ for $t\in(0,\pi)$ and $R\rightarrow \infty$ vanishes and can be added without changing the result.

Finally, the obtained closed contour can be shrinked around the cut mentioned above. Change of variables $z=i(1+t)$ should give you a representation of the beta-function.

Diger
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