My question is the following

Question. Can you compute some of the following $$c_{n,k}=\int_0^{\pi/2}(\sin x)^n e^{-(2+\cos x)\log k}dx$$ where $n\geq 1$ is a fixed integer and $k\geq 1$ is also a fixed integer? What of those? Thanks in advance.

If you know some of related integrals you can add also a link from this site of those computations that provide closed forms for our integrals in previous Question. For the other, you can add yourself computations/reductions.

My attempt was convice to me that seems a hard problem, because for the lower odds $n=1$ or $n=3$ using the rule $$\int F'(x)e^{F(x)}dx=e^{F(x)}+\text{constant}$$ and integration by part, is a lot of work yet. I take the case $k=1$ as a special case, thus I am knowing that from Wolfram Alpha $\int_0^{\pi/2}(\sin x)^ndx$ a closed form is provide us with the code

$$\int_0^{\pi/2} \sin^n(x)\, dx$$

for integers $n\geq 1$. But the code for a reduction with $n=6$, this is

$$\int_0^{\pi/2} \sin^6(x)\,e^{-\cos(x)}\, dx$$

doesn't provide to me a closed-form. For example $n=5$ likes to me tedious computations.

The context of my computations is speculative, I don't know if there were mistakes or my reasoning/approach is reasonable (please, feel free to add comments if you detect some): let the complex variable $s=\sigma+it$ , then $\zeta(s)$ the Riemann Zeta function is analytic for $\Re s>1$ and doen't vanishes. Is known that $$\frac{1}{\zeta(s)}=\sum_{n=1}^\infty\frac{\mu(n)}{n^s},$$ for $\Re s>1$. I believe that I can define the following real functions, as restrictions ($X=x+0\cdot i$) of previous complex function, $$f(X)=\frac{1}{\zeta(X+2)}$$ as continuous for $-1<X\leq 1$, thus this next composition $f: \left[ 0,1\right] \to\mathbb{R}$, also as continuous real function $$f(\cos x)=\frac{1}{\zeta(2+\cos x)}=\sum_{k=1}^\infty\mu(k)k^{-2-\cos x},$$ when (if I can presume it) $$-1<\cos x\leq 1.$$

Then as a specialization of the PROBLEMA 159 in page 515, in spanish, proposed by Furdui in La Gaceta de la RSME Vol. 14 (2011) No. 3, I say the first statement, and the first line of the Solution that provide us
$$\frac{\sqrt{\frac{\pi}{2}}}{\zeta(3)}=\lim _{n\to\infty}\sqrt{n}\sum_{k=1}^\infty\mu(k)c_{n,k},$$ when is justified to swap the sign of series $\sum_{k=1}^\infty$ with the definite integral.

  • $\begingroup$ Feel free if you want to add a new tag. Thanks. $\endgroup$ – user243301 Jun 8 '16 at 10:36
  • $\begingroup$ Very thanks much to Leibovici and user1952009, for their approach and remarks. Now I am studying their solutions and remarks. Very thanks much all users by their patience and attention. $\endgroup$ – user243301 Jun 8 '16 at 11:20
  • $\begingroup$ All users, when I am claiming about Wolfram Alpha, is always with respect its online calculator with a standard computation time and myself code. Thus I am assuming that with more time it provide us the best results. $\endgroup$ – user243301 Jun 8 '16 at 12:30

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} \color{#f00}{c_{n,k}} & \equiv \int_{0}^{\pi/2}\sin^{n}\pars{x}\expo{-\bracks{2 + \cos\pars{x}}\ln\pars{k}} \,\dd x = {1 \over k^{2}}\int_0^{\pi/2}\sin^{n}\pars{x}\,k^{-\cos\pars{x}}\,\dd x \\[3mm] & = -\,{1 \over k^{2}}\int_{x = 0}^{x = \pi/2} \bracks{1 - \cos^{2}\pars{x}}^{\pars{n - 1}/2}\, k^{-\cos\pars{x}}\,\dd\bracks{\cos\pars{x}} \\[3mm] & \stackrel{\cos\pars{x}\ \to\ x}{=}\ {1 \over k^{2}}\int_{0}^{1}\pars{1 - x^{2}}^{\pars{n - 1}/2}\,k^{-x}\,\dd x = {1 \over k^{2}}\int_{0}^{1}\pars{1 - x^{2}}^{\pars{n - 1}/2}\,\expo{-\mu x} \,\dd x \\[3mm] &\ \mbox{where}\ \mu \equiv \ln\pars{k} \end{align}

Then, \begin{align} \color{#f00}{c_{n,k}} & = {1 \over k^{2}}\bracks{% -\,{\root{\pi}\Gamma\pars{n/2 + 1/2} \over 2\pars{\mu/2}^{n/2}}\, \mathbf{\mathrm{M}}_{n/2}\pars{\mu}} \\[3mm] & = \color{#f00}{-\,{\root{\pi} \over k^{2}}2^{n/2 - 1}\ln^{-n/2}\pars{k}\, \Gamma\pars{{n \over 2} + \half}\mathbf{\mathrm{M}}_{n/2}\pars{\ln\pars{k}}}\,, \qquad \Re\pars{n} > -1 \end{align}

$\mathbf{\mathrm{M}}_{\nu}\pars{z}$ is the Modified Struve Function. See, for example, $\mathbf{11.5.4}$ ( page 292 ) in NIST HandBook of Mathematical Functions or/and DLMF-NIST page.

  • $\begingroup$ Very thanks much, tomorrow I am studying the details of your proof. $\endgroup$ – user243301 Jun 12 '16 at 11:53
  • $\begingroup$ @user243301 Thanks. You are welcome. $\endgroup$ – Felix Marin Jun 12 '16 at 21:21
  • $\begingroup$ Very thanks much to you and Leibovici for your proofs to get this coefficients. I am agree that (on assumption that my previous computations were rights) that this expression for this constant is very complicated, but now your computations and @ClaudeLeibovici can be as example of the compution of this kind of integrals. Very thanks much then for your and Leibovici computations that now are a reference for all users. $\endgroup$ – user243301 Jun 13 '16 at 10:36
  • $\begingroup$ @user243301 Thanks. You are always welcome. Claude Leibovici always posts very fine answers. $\endgroup$ – Felix Marin Jun 13 '16 at 23:03

This is not an answer since the result comes from a CAS.

$$\int_{0}^{\frac{\pi}2}\sin^n(x)\,dx=\frac{\sqrt{\pi } \Gamma \left(\frac{n+1}{2}\right)}{2 \Gamma \left(\frac{n}{2}+1\right)}$$ $$\int_{0}^{\frac{\pi}2}\sin^n(x)e^{-\cos(x)}\,dx=\sqrt{\pi }\, 2^{\frac{n}{2}-1} \,\Gamma \left(\frac{n+1}{2}\right) \left(I_{\frac{n}{2}}(1)-\pmb{L}_{\frac{n}{2}}(1)\right)$$ where appear Bessal and Struve functions. Both are of $\Re(n)>-1$.

It evens seems that $$\int_{0}^{\frac{\pi}2}\sin^n(x)e^{-a\,\cos(x)}\,dx=\sqrt{\pi }\, 2^{\frac{n}{2}-1}\, a^{-n/2}\, \Gamma \left(\frac{n+1}{2}\right) \left(I_{\frac{n}{2}}(a)-\pmb{L}_{\frac{n}{2}}(a)\right)$$

May be (I hope), this will give you some ideas.

  • 1
    $\begingroup$ the Bessel function appears when applying the residue theorem to $\int_0^{2 \pi} \sin^n(x) e^{a \cos(x)} dx = \int_{|z| = 1} (\frac{z-z^{-1}}{2i})^n e^{a \frac{z+z^{-1}}{2}} \frac{dz}{i z}$ with the (complex) change of variable $z= e^{it}, dz = i e^{it} dt, dt = \frac{dz}{i z}$ $\endgroup$ – reuns Jun 8 '16 at 10:50
  • $\begingroup$ @user1952009. I am terribly poor with residues. Thanks for pointing it. $\endgroup$ – Claude Leibovici Jun 8 '16 at 10:52

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