Integrate $I=\int_0^{\infty} x^n \, e^{ax+\frac{b}{x}} \, \cos(cx) \, dx$?

Question

Is there an expression in terms of some special functions (or a closed form) of the following integral

$$I_n(a,b,c)=\int_0^{\infty} x^n \, e^{ax+\frac{b}{x}} \, \cos(cx) \, dx,$$ $n:$ integer,

$a\in\mathbb R; \, a<0$,

$b\in \mathbb C; \, \Re e\, b<0$

and $c\in \mathbb R$.

Remark: In [I. S. Gradshteyn, I. M. Ryzhik, Table of Integrals, Series and Products, New York, Academic, 1980. p:486], there exists a closed form of the following integral

$$I(\alpha,\beta,a)=\int_0^{\infty} \, e^{-\alpha x^2-\frac{\beta}{x^2}} \, \cos(ax^2) \, dx, \quad \Re e\, \alpha>0, \Re e\, \beta>0$$ given by $$C e^{-2c\sqrt{\beta}} \left[C_1\, \cos(C_3\sqrt{\beta})- C_2\, \sin(C_3\sqrt{\beta})\right],$$ where $C, C_1, C_2, C_3$ constants depend on $\alpha$ and $\beta$.

Thank you in advance

Answer 1

If we define: $$I_1=\frac 12\int_0^\infty x^ne^{-b/x}e^{-(a+ic)x}dx$$ $$I_2=\frac 12\int_0^\infty x^ne^{-b/x}e^{-(a-ic)x}dx$$ then $I=I_1+I_2$ and differentiating we get: $$dI_1=nI_1(n-1)dn-I_1(n+1)da-I_1(n-1)db-iI_1(n+1)dc$$ $$dI_2=nI_2(n-1)dn-I_2(n+1)da-I_2(n-1)db+iI_2(n+1)dc$$ so maybe you can find some kind of recurrance relation?

Answer 2

A slight modification to the OP's definition: $$\tilde{I}_n(a,b,c):=\int_0^\infty x^n\,\exp{\big(-(ax+b/x)\big)}\cos(cx) \,dx \quad a,b,c>0 $$ Then by differentiating with respect to $a,$ $$\tilde{I}_n(a,b,c):=(-1)^n \,\frac{d^n}{da^n} \int_0^\infty \exp{\big(-(ax+b/x)\big)}\cos(cx) \,dx \,.$$ Use an exponential for the cosine, $$\tilde{I}_n(a,b,c):=(-1)^n \,\frac{d^n}{da^n} Re\Big[\int_0^\infty \exp{\big(-((a+ic)\,x+b/x)\big)} \,dx \Big]\,.$$ The integral is solvable in closed form in terms of MacDonald (a variant of Bessel) functions. $$\tilde{I}_n(a,b,c):=(-1)^n \,\frac{d^n}{da^n} Re\Big[ 2\sqrt{b} \,\, \frac{K_1 \big(2 \sqrt{b}\sqrt{a+ic}\big)}{\sqrt{a+ic}} \Big]$$

There are some nice things about this representation. It's easily coded on Matematica. Asymptotics are well-known, if some of the parameters get large. The derivatives of Bessel functions give higher order Bessel funtions, but recursion will always take the answer to a polynomial times a $K_0$ and a $K_1.$ For example,

$$\tilde{I}_2(a,b,c)=Re\Big[2\sqrt{b} \,\, \frac{K_0 \big(2 \sqrt{b}\sqrt{a+ic}\big)}{(a+ic)^2} + (2+b(a+ic)) \,\, \frac{K_1 \big(2 \sqrt{b}\sqrt{a+ic}\big)}{(a+ic)^{5/2}} \Big] $$

Answer 3

Well, we have the following integral:

$$\mathcal{I}_\text{n}\left(\alpha,\beta,\gamma\right):=\int_0^\infty x^\text{n}\exp\left(\alpha x+\frac{\beta}{x}\right)\cos\left(\gamma x\right)\space\text{d}x=$$ $$\int_0^\infty x^\text{n}\exp\left(\alpha x\right)\cos\left(\gamma x\right)\exp\left(\frac{\beta}{x}\right)\space\text{d}x\tag1$$

Using the evaluating integrals over the positive real axis property of the Laplace transform, we can write:

$$\mathcal{I}_\text{n}\left(\alpha,\beta,\gamma\right)=\int_0^\infty\mathcal{L}_x\left[x^\text{n}\exp\left(\alpha x\right)\cos\left(\gamma x\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[\exp\left(\frac{\beta}{x}\right)\right]_{\left(\text{s}\right)}\space\text{ds}\tag2$$

Using properties of the Laplace transform, we can write:

$$\mathcal{I}_\text{n}\left(\alpha,\beta,\gamma\right)=\int_0^\infty\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathcal{L}_x\left[\exp\left(\alpha x\right)\cos\left(\gamma x\right)\right]_{\left(\text{s}\right)}\right)\cdot\mathcal{L}_x^{-1}\left[\exp\left(\frac{\beta}{x}\right)\right]_{\left(\text{s}\right)}\space\text{ds}=$$ $$\int_0^\infty\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathcal{L}_x\left[\cos\left(\gamma x\right)\right]_{\left(\text{s}-\alpha\right)}\right)\cdot\mathcal{L}_x^{-1}\left[\exp\left(\frac{\beta}{x}\right)\right]_{\left(\text{s}\right)}\space\text{ds}\tag3$$

Using the table of selected Laplace transforms, we can write:

$$\mathcal{I}_\text{n}\left(\alpha,\beta,\gamma\right)=\int_0^\infty\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{\text{s}-\alpha}{\left(\text{s}-\alpha\right)^2+\gamma^2}\right)\cdot\mathcal{L}_x^{-1}\left[\exp\left(\frac{\beta}{x}\right)\right]_{\left(\text{s}\right)}\space\text{ds}\tag4$$

Using:

$$\exp(x)=\sum_{\text{k}\ge0}\frac{x^\text{k}}{\text{k}!}\tag5$$

Where $\exp(\cdot)$ is the Exponential function.

So, we can write:

$$\mathcal{I}_\text{n}\left(\alpha,\beta,\gamma\right)=\int_0^\infty\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{\text{s}-\alpha}{\left(\text{s}-\alpha\right)^2+\gamma^2}\right)\cdot\mathcal{L}_x^{-1}\left[\sum_{\text{k}\ge0}\frac{\left(\frac{\beta}{x}\right)^\text{k}}{\text{k}!}\right]_{\left(\text{s}\right)}\space\text{ds}=$$ $$\int_0^\infty\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{\text{s}-\alpha}{\left(\text{s}-\alpha\right)^2+\gamma^2}\right)\cdot\sum_{\text{k}\ge0}\frac{\beta^\text{k}}{\text{k}!}\cdot\mathcal{L}_x^{-1}\left[\frac{1}{x^\text{k}}\right]_{\left(\text{s}\right)}\space\text{ds}\tag6$$

Using the table again, we get:

$$\mathcal{I}_\text{n}\left(\alpha,\beta,\gamma\right)=\int_0^\infty\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{\text{s}-\alpha}{\left(\text{s}-\alpha\right)^2+\gamma^2}\right)\cdot\sum_{\text{k}\ge0}\frac{\beta^\text{k}}{\text{k}!}\cdot\frac{\text{s}^{\text{k}-1}}{\Gamma\left(\text{k}\right)}\space\text{ds}=$$ $$\sum_{\text{k}\ge0}\frac{\beta^\text{k}}{\text{k}!}\cdot\frac{\left(-1\right)^\text{n}}{\Gamma\left(\text{k}\right)}\int_0^\infty\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{\text{s}-\alpha}{\left(\text{s}-\alpha\right)^2+\gamma^2}\right)\cdot\text{s}^{\text{k}-1}\space\text{ds}\tag7$$