Finite integral with goniometric functions, $\int_0^{\infty} \frac{8\sin^4(\pi f t)\tan^2(\pi f/2)}{(\pi^4 \tau^2 f^3) }df$ I have difficulties trying to find an algebraic solutions of the following integral:

The $\tau$ in this formula is an integer, which is a very important fact because only then this integral is convergent. I want to find a solution as a function of $\tau$. Using Mathematica I can only find algabraic solutions when I fill in a certain $\tau$. So I think there is an algebraic solutions for all $\tau$ but I cannot find it yet. This is a list of answers for $\tau=1 $ to $7$:

If anyone knows how to deal with this problem, I am very pleased to hear it.
Chantal
 A: Here is an idea. I have not made all the computations, and I hope that I does not  have  made computation errors.
If you make the change of variable $2t=f\pi$, then your integral becomes
$$C\int_0^{+\infty}\frac{(\sin(2t\tau))^4\sin(t)^2}{t^3\cos(t)^2} dt=C I$$
where $C$ is a computable constant. Now as $\tau$ is an integer, there exists a polynomial $A_{\tau}\in \mathbb {R}[x]$ such that $\displaystyle \sin(2\tau t)=\sin(t)A_{\tau}(\cos(t))$. Putting $t=\pi/2$, we see that $A_{\tau}(0)=0$, hence $A_{\tau}(x)=xB_{\tau}(x)$. Now
$$I=\int_0^{+\infty}\frac{(\sin(t))^6\cos(t)^2(B_{\tau}(\cos(t))^4}{t^3}dt$$
We can write
$$((\sin(t))^6\cos(t)^2(B_{\tau}(\cos(t))^4=\sum_{m=0}^N b_m \cos(mt)$$
Let $\varepsilon>0$, and $$I_{\varepsilon}=\int_{\varepsilon}^{+\infty}\frac{((\sin(t))^6\cos(t)^2(B_{\tau}(\cos(t))^4}{t^3}dt=\sum_{m=0}^N b_m\int_{\varepsilon}^{+\infty}\frac{\cos(mt)}{t^3} dt$$
Of course, $I_{\varepsilon}\to I$ as $\varepsilon\to 0$.
Now for $m\geq 1$:
$$\int_{\varepsilon}^{+\infty}\frac{\cos(mt)}{t^3}dt=m^2\int_{m\varepsilon}^{+\infty}\frac{\cos(t)}{t^3}dt$$
We have:
$$\int_{\varepsilon}^{+\infty}\frac{\cos(t)}{t^3}dt=\int_{\varepsilon}^1 \frac{\cos(t)-1+t^2/2}{t^3}dt+\int_1^{+\infty}\frac{\cos(t)}{t^3}dt-\frac{1}{2}+\frac{1}{2\varepsilon^2}+\frac{1}{2}\log \varepsilon$$ or
$$\int_{\varepsilon}^{+\infty}\frac{\cos(t)}{t^3}dt=J(\varepsilon)+K-\frac{1}{2}+\frac{1}{2\varepsilon^2}+\frac{1}{2}\log \varepsilon$$
Put $\displaystyle \alpha=\sum_{m=1}^N b_m$, $\displaystyle \beta=\sum_{m=1}^N b_m m^2\log m$, $\displaystyle \gamma=\sum_{m=1}^N b_m m^2$. We have:
$$I_{\varepsilon}=\frac{1}{2\varepsilon^2}(b_0+\alpha)+\frac{\beta}{2}+\frac{\gamma}{2}\log \varepsilon+\sum_{m=1}^N b_m m^2J(m\varepsilon)+(K-\frac{1}{2})\gamma$$
Now we immediately see that $\displaystyle J(\varepsilon)\to J=\int_{0}^1 \frac{\cos(t)-1+t^2/2}{t^3}dt$ if $\varepsilon \to 0$. Hence we must have $b_0+\alpha$=0 (multiply by $\varepsilon$, and $\varepsilon \to 0$), and then that $\gamma=0$. Hence:
$$2I=\sum_{m=1}^N b_m m^2\log m$$
It remains to compute the $b_m$....
