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I would like to solve the following integral that is a variation of this one (Integral involving Modified Bessel Function of the First Kind).

Namely, I have:

$$\frac{1}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{(-\frac{x}{\alpha}-\frac{1}{2w^2}(x-\hat{x})^2)} \, I_0\left(\frac{x}{\beta}\right)\,dx .$$

By using the series representation for $I_0(x)$, I obtain the Tricomi hypergeometric function and I don't know how to calculate the corresponding series.

Any suggestion is highly appreciated.

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  • $\begingroup$ I edited your post. Check if any change happened. $\endgroup$ – Mhenni Benghorbal Mar 18 '13 at 14:04
  • $\begingroup$ Why don't you post the result you have found, so it make it easier for those who would assist you. $\endgroup$ – Mhenni Benghorbal Mar 18 '13 at 14:49
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One can try to do the following derivations. $$\mathrm{Int}=\frac{1}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{(-\frac{x}{\alpha}-\frac{1}{2w^2}(x-\hat{x})^2)} \, I_0\left(\frac{x}{\beta}\right)\,dx$$ You can simplify the power of the exponent: $$-\left(\frac{x}{\alpha}+\frac{1}{2w^2}(x-\hat{x})^2\right)=-\left(\frac{(x-\mu)^2}{2w^2}+\gamma \right),$$ where $\gamma=\left(\frac{\hat{x}}{\alpha}-\frac{w^2}{2\alpha^2}\right)$, $\mu=\hat{x}-\frac{w^2}{\alpha^2}$. So: $$\mathrm{Int}=\frac{1}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{(-\frac{x}{\alpha}-\frac{1}{2w^2}(x-\hat{x})^2)} \, I_0\left(\frac{x}{\beta}\right)\,dx=\frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{-\frac{(x-\mu)^2}{2w^2}} \, I_0\left(\frac{x}{\beta}\right)\,dx$$ Then you can change the variable of integration $y=x-\mu$: $$\mathrm{Int}=\frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, (y+\mu) \, e^{-\frac{y^2}{2w^2}} \, I_0\left(\frac{y+\mu}{\beta}\right)\,dx$$ and make use of the Neumann’s addition theorem: $$\mathop{I_{{\nu}}}\nolimits\!\left(u\pm v\right)=\sum _{{k=-\infty}}^{\infty}(\pm 1)^{k}\mathop{I_{{\nu+k}}}\nolimits\!\left(u\right)\mathop{I_{{k}}}\nolimits\!\left(v\right)$$ setting $\nu=0$, assuming that $y,\mu,\beta\in\mathbb{R}$ and using connection formulas one will get: $$\mathop{I_{0}}\nolimits\!\left(\frac{y+\mu}{\beta}\right)=\sum _{{k=-\infty}}^{\infty}\mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{k}}}\nolimits\!\left(\frac{\mu}{\beta}\right)=\mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{0}}}\nolimits\!\left(\frac{\mu}{\beta}\right)+2\sum _{{k=1}}^{\infty}\mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{k}}}\nolimits\!\left(\frac{\mu}{\beta}\right)$$

$$\mathrm{Int}=\! \frac{\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)e^{-\gamma}}{\sqrt{2\pi w^2}}\!\int_{-\infty}^{+\infty} \, (y+\mu)\! e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx\! +\!\frac{2 e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty}\! (y+\mu) \, e^{-\frac{y^2}{2w^2}} \! \sum _{k=1}^{\infty}\!\mathop{I_k}\!\!\left(\frac{y}{\beta}\right)\!\!\mathop{I_k}\nolimits\!\left(\frac{\mu}{\beta}\right) \,dx$$ Or reorganising terms: $\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)$

$$\mathrm{Int}=\! \frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\!\left(\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)\int_{-\infty}^{+\infty} \, (y\!+\!\mu)\! e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx\! +\!2\!\sum _{k=1}^{\infty}\!\mathop{I_k}\!\left(\frac{\mu}{\beta}\right)\!\int_{-\infty}^{+\infty}\! (y\!+\!\mu) \, e^{-\frac{y^2}{2w^2}} \! \!\mathop{I_k}\!\left(\frac{y}{\beta}\right)\!\!\,dx \right)$$ Let's set $$\mathrm{Int}_k(\mu,\alpha,\beta)\!=\!\int_{-\infty}^{+\infty} \, (y\!+\!\mu) e^{-\frac{y^2}{2w^2}}\! \mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx$$ $$\mathrm{Int}_0(\mu,\alpha,\beta)\!=\!\int_{-\infty}^{+\infty} \, (y\!+\!\mu) e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx$$ Here one can play with the evenness/unevenness of the functions under integrals (as I did at first), or just use CAS and get:

$$\mathrm{Int}_k(\mu,\alpha,\beta)\!=\!\sqrt{\frac{\pi \alpha ^2}{8 \beta ^2}}\! e^{\frac{\alpha ^2}{4 \beta ^2}} \left(\!\!2 \beta \left(\!\!(-1)^k+1\!\right)\! \mu I_{\frac{k}{2}}\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)+\alpha ^2 \left(\!1-(-1)^k\!\right)\!\! \left(\!\!I_{\frac{k-1}{2}}\!\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)+I_{\frac{k+1}{2}}\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)\!\!\right)\!\!\right)$$ $$\mathrm{Int}_0(\mu,\alpha,\beta)\!=\! \sqrt{2 \pi } \alpha \mu e^{\frac{\alpha ^2}{4 \beta ^2}} I_0\left(\frac{\alpha ^2}{4 \beta ^2}\right)$$ So the original integral can be represented in the following way: $$\mathrm{Int}(\mu,\alpha,\beta)\!=\! \frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\left(\mathrm{Int}_0(\mu,\alpha,\beta)+ 2\!\sum _{k=1}^{\infty}\mathop{I_k}\left(\frac{\mu}{\beta}\right) \mathrm{Int}_k(\mu,\alpha,\beta) \right)$$ And I am not shure that there is any chance to simplify it futher (meaning finding the sum of the series).

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I'm getting rid of all the unnecessary parameters (and setting $x \to \beta x$):

$$\int_{-\infty}^{\infty} x ~ e^{-a x- b^2 (x-c)^2} I_0 (x) dx$$

We'll also solve an easier integral:

$$J(a,b,c)=\int_{-\infty}^{\infty}e^{-a x- b^2 (x-c)^2} I_0 (x) dx$$

To get back to the first integal we only need to take $a$ derivative:

$$ \frac{\partial J(a,b,c)}{\partial a}=-\int_{-\infty}^{\infty} x ~ e^{-a x- b^2 (x-c)^2} I_0 (x) dx$$

Now we use the integral representation of the Bessel function:

$$I_0(x)=\frac{1}{\pi} \int_0^{\pi} e^{x \cos t} dt$$

Then we have:

$$J=\frac{1}{\pi} \int_0^{\pi} \int_{-\infty}^{\infty} e^{-(a-\cos t) x- b^2 (x-c)^2}~ dx~ dt$$


Let's complete the square (setting $a-\cos t=\gamma(t)$ for now):

$$-\gamma x- b^2 (x-c)^2=-b^2 \left(x-c+\frac{\gamma}{2 b^2} \right)^2+\frac{\gamma^2}{4 b^2}-c \gamma$$

So the integral becomes:

$$J=\frac{1}{\pi} \int_0^{\pi} \exp \left( \frac{\gamma^2}{4 b^2}-c \gamma \right) \int_{-\infty}^{\infty} e^{-b^2 z^2}~ dz~ dt=$$

$$=\frac{1}{\sqrt{ \pi}~ b} \int_0^{\pi} \exp \left( \frac{\gamma(t)^2}{4 b^2}-c \gamma(t) \right) ~dt$$

Getting back to the original expression for $\gamma$:

$$\frac{\gamma(t)^2}{4 b^2}-c \gamma(t)=\frac{a^2}{4 b^2}-ac+\left(c-\frac{a}{2 b^2} \right) \cos t+\frac{1}{4 b^2} \cos^2 t$$

$$J=\frac{1}{\sqrt{ \pi}~ b} \exp \left( \frac{a^2}{4 b^2}-ac \right) \int_0^{\pi} \exp \left( \left(c-\frac{a}{2 b^2} \right) \cos t+\frac{1}{4 b^2} \cos^2 t \right) ~dt$$

Now I'll rename the parameters again for clarity:

$$J=\frac{A}{\pi} \int_0^{\pi} e^{ B \cos t+C \cos^2 t } ~dt$$


And now there is only series solution available (as far as I know). Expanding the second exponent:

$$ e^{C \cos^2 t }=1+C \cos^2 t+\frac{C^2}{2} \cos^4 t+\cdots+\frac{C^{k}}{k!} \cos^{2k} t+\cdots$$

$$I_0(B)= \frac{1}{\pi} \int_0^{\pi} e^{ B \cos t} ~dt $$

$$\frac{d^2 I_0(B)}{d B^2}= \frac{1}{\pi} \int_0^{\pi} e^{ B \cos t} \cos^2 t~dt =\frac{1}{2} (I_0(B)+I_2(B))$$

And so on. We obtain a series:

$$J=A \sum_{n=0}^{\infty} \frac{C^n}{n!} I_0^{(2n)} (B)$$

Here $I_0^{(2n)}$ - $2n^{th}$ derivative of $I_0(B)$ w.r.t. $B$.

$$A=\frac{\sqrt{\pi}}{b} \exp \left( \frac{a^2}{4 b^2}-ac \right)$$

$$B=c-\frac{a}{2 b^2} $$

$$C=\frac{1}{4 b^2}$$

I'll leave it for you to recover your original parameters and find:

$$\int_{-\infty}^{\infty} x ~ e^{-a x- b^2 (x-c)^2} I_0 (x) dx=-\frac{\partial}{\partial a} A \sum_{n=0}^{\infty} \frac{C^n}{n!} I_0^{(2n)} (B)$$

$$\int_{-\infty}^{\infty} x ~ e^{-a x- b^2 (x-c)^2} I_0 (x) dx=A \left(B \sum_{n=0}^{\infty} \frac{C^n}{n!} I_0^{(2n)} (B)+\frac{C}{2} \sum_{n=0}^{\infty} \frac{C^n}{n!} I_0^{(2n+1)} (B) \right)$$

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