Indefinite integral involving product of sinc with cosine: $\displaystyle\int\operatorname{sinc}^2(ax) \cos(bx)dx$

Question

I try to solve this indefinite integral:

\begin{equation} \int\,\mathrm{sinc}^{2}\left(ax\right)\,\cos\left(bx\right)\,\,\mathrm{d}x \end{equation}

whith $\mathrm{sinc}\left(x\right) = \frac{\sin\left(x\right)}{x}$. I tried to do integratin by parts using the D-I-method / tabular method. However, this resulted in very lengthy and complicated expressions, independent which part is considered for differentiation and for integration. Do you know some other strategy how to obtain a solution? Of course I checked WolframAlpha and seeing that there exists a solution lets me feel at least certain that an analytic solution exists.

https://www.wolframalpha.com/input?i=integrate+sinc%28a*x%29%5E%282%29+*+cos%28b*x%29+dx

Your help is appreciated, thank you :)

Answer 1

We first rewrite the integral $$\int \operatorname{sinc}^2 ax \cos bx \,dx$$ (for $a \neq 0$) in terms of $\sin$ and $\cos$, giving a constant multiple of $$I_{a, b} := \int \frac{\sin^2 a x \cos b x \,dx}{x^2} .$$ Applying a double-angle identity and then a product-to-sum identity then yields \begin{multline}I_{a, b} = \int \frac{\frac12 (1 - \cos 2 a x) \cos b x \,dx}{x^2} \\= \frac{1}{2} \left[\int \frac{\cos b x \,dx}{x^2} - \frac12 \int \frac{\cos (2 a + b) x + \cos (2 a - b) x}{x^2} \,dx\right] .\end{multline} Then, in terms of $$J_k := \int \frac{\cos k x \,dx}{x^2},$$ the integral $I_{a, b}$ is $$I_{a, b} = \frac12 J_b - \frac14 J_{2 a + b} - \frac14 J_{2 a - b} .$$ Now, to evaluate $J_k$, we integrate by parts with $u = \cos k x$, $v = \frac{dx}{x^2}$, giving $$J_k = - \frac{\cos k x}{x} - k \int \frac{\sin k x}{x} \,dx = -\frac{\cos k x}{x} - k \operatorname{Si}(k x) + C,$$ where $\operatorname{Si}(u) := \int_0^u \frac{\sin t \,dt}{t}$ is the sine integral function.

Putting everything together and reorganizing the $\cos$ terms for compactness gives \begin{multline} \color{#bf0000}{ \boxed{ \small \int \operatorname{sinc}^2 ax \cos bx \,dx \\ \small = \frac{1}{4 a^2}\left[(2 a - b) \operatorname{Si} ((2 a - b) x) - 2 b \operatorname{Si} (b x) + (2 a + b) \operatorname{Si} ((2 a + b) x)\right] \small - \frac{\sin^2 ax \cos b x}{a^2 x} + C}} . \end{multline}

The case $b = 0$ reduces to $$\int \operatorname{sinc}^2 u \,du = \operatorname{Si}(2 u) - \frac{\sin^2 u}{u} + C .$$

Answer 2

How about a reduction formula?
(I took the case $a=b=1$ of the original question.)
Consider integrals of the form $$ J(a,b,c) = \int\frac{\sin^a(x)\cos^b(x)}{x^c}\;dx , $$ where $c$ is a positive integer. We want a "reduction formula" to evalute this in terms of similar integrals with smaller value of $c$.
Integrate by parts to get something like $$ J(a,b,c) = -\frac{\sin^a(c)\cos^b(x)}{(c-1)x^{c-1}} +\frac{1}{c-1}\int\left(\frac{(a+b)\sin^{a-1}(x)\cos^{b+1}(x)}{x^{c-1}} -\frac{b\sin^{a-1}(x)\cos^{b-1}(x)}{x^{c-1}}\right)\;dx \\ =-\frac{\sin^a(c)\cos^b(x)}{(c-1)x^{c-1}} +\frac{a+b}{c-1}J(a-1,b+1,c-1)-\frac{b}{c-1}J(a-1,b-1,c-1) . $$ Starting values $J(a,b,1)$ involve sine integral $\operatorname{Si}$ and/or cosine integral $\operatorname{Ci}$.