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I have the following definite integral:

$$ \int_{0}^{\pi} cos(n\theta)[I_{1}(acos\theta) + cos\theta I_{0}(acos\theta)]d\theta $$

with $I_{1}$ and $I_{0}$ the first and zero-order modified bessel functions of the first kind respectively ; $n \in \mathbb{N}^{*}$ and $a \in \mathbb{R}^{+*}$.

with the hint that,

$$ I_{n}(z) = \frac{1}{\pi} \int_{0}^{\pi} e^{zcos\theta}cos(n\theta)d\theta $$

we can transform the previous integral,

$$ \int_{0}^{\pi} cos(\theta)I_{n}(acos\theta) + \frac{1}{2}(I_{n+1}(acos\theta) + I_{n-1}(acos\theta))d\theta $$ $$ \int_{0}^{\pi} cos(\theta)I_{n}(acos\theta) - \frac{1}{asin\theta}(I_{n}^{'}(acos\theta))d\theta $$

I couldn't go any further in the computation. But using the power series representation and considering $n$ odd and $n=2m+1$, we obtain:

$$ \pi \sum_{k=0}^{\infty} \frac{U^{2k+n-1}}{2^{2k+n}\Gamma (k+n+1) k!} \prod_{i=1}^{k+m} (\frac{2i-1}{2i}) (U\frac{2(k+m)+1}{2(k+m+1)} + 2k+n) $$

If someone has an idea ... Thanks!

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    Basically, you're trying to find the cosine transform of $I_{1}(a\cos\theta) + I_{0}(a\cos\theta)\cos\theta$, then? – 2017-01-31
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    Yes, actually, it can be written as the integral of $cos(n\theta)(I_{1}(acos\theta) + I_{0}(acos\theta)cos\theta)d\theta$. – 2017-01-31
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    Do you *really* need the indefinite integral? The definite integral over $(-1,1)$ asks for the Fourier-Chebyshev series of $I_1(x)$, i.e. for the Fourier series of $I_1(\cos x)$, that is straightforward given the definition of $I_1$ as a coefficient of a Fourier series. – 2017-01-31
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    @JackD'Aurizio You are right, if I understand it well, we can consider the $2\pi$-periodic function $s(\theta) = I_{1}(acos\theta) + I_{0}(acos\theta)cos\theta$ and the definite integral $a_{n} = \frac{1}{\pi} \int_{0}^{2\pi}cos(n\theta)s(\theta)$ is a coefficient of a Fourier series. But how can I evaluate this coefficient ? – 2017-02-01
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    @LA.Dufrene: $$I_n(z)=\frac{1}{\pi}\int_{0}^{\pi}e^{z\cos\theta}\cos(n\theta)\,d\theta$$ is the key relation, if your problem deals with the integral over $(-1,1)$. Is it so? – 2017-02-01
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    @JackD'Aurizio: After checking my problem, yes, the integral of the original post (with variable x) can also be defined on $(-1,1)$ or equivalently on $(0,\pi)$ with variable $\theta$. – 2017-02-02
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    @JackD'Aurizio: Thanks Jack, but I still don't understand how to use this relation to solve my problem. – 2017-02-02
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    @JackD'Aurizio: Thanks Jack for your help, but I couldn't find the answer, so I'm giving up now. I'll try to find a Maple license. Eventually, if I find the answer I will post it in comment. – 2017-02-06
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    @JackD'Aurizio: Jack you seemed to know the answer, could you confirm the edit and eventually giving me another hint ? – 2017-02-17

1 Answers 1

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Take the integral

$$ \int_{0}^{\pi} cos(\theta) I_{n}(acos(\theta)) + \frac{1}{2}(I_{n+1}(acos(\theta)) + I_{n-1}(acos(\theta))) d\theta $$

and apply the solution of the integral here, p724, 6.681-3:

$$ \int_{0}^{\frac{\pi}{2}} cos(2\mu x)I_{2\nu}(2acosx)dx = \frac{\pi}{2}I_{\nu-\mu}(a)I_{\nu+\mu}(a) $$

This is also valid for the sinus by symmetry in our case. Stating $n=2m+1$ we obtain,

$$ \pi I_{m}(\frac{a}{2})I_{m+1}(\frac{a}{2}) + \frac{\pi}{2}I_{m+1}(\frac{a}{2})^{2} + \frac{\pi}{2}I_{m}(\frac{a}{2})^{2} $$