# Integral concerning $\sin^2$

Consider the following integral :

$$I(x;a)= \int \frac{\sin^2(x+1)}{(x+1)^a}\,dx$$

Now let the $$x$$ runs from $$0$$ to $$\infty$$

i.e. let

$$I_p(a)= \int_0^\infty \frac{\sin^2(x+1)}{(x+1)^a}\,dx$$

Now we can see that $$I_p(a)$$ converges for all $$\text{R}(a)>1$$.

Also, for $$a=1$$ , $$I_p(a)$$ diverges .

Question 1 : If possible , can $$I_p(a)$$ be represented in following way :

$$I_p(a) = f(a) + \frac{C}{(a-1)}$$

Here , $$f(a)$$ is convergent for all $$a\geqslant1$$ and $$C>0$$

The factor $$\frac{C}{(a-1)}$$ explains the divergence of $$I_p(a)$$ at $$a=1$$

Question 2 ( important ):

Analogous to Question 1 : Only replace $$\sin^2(x+1)$$ by $$\sin^2(g(x))$$ in $$I_p(a)$$

where, $$g(x)$$ is an increasing function in $$[1,\infty)$$

• @JackD'Aurizio: why doesn't $I(x;a)$ depend on $x$? Isn't it the antiderivative of a function of $x$? Commented Sep 27, 2019 at 8:28
• Someone downvoted ? Is something wrong ? Commented Sep 27, 2019 at 8:37
• Pardon the bad wording, I simply meant that $I(x,a)$ is related to the sine integral, which is known to be non-elementary. $I_p(a)$ can be written as $\int_{1}^{+\infty}\sin(x)/x^a\,dx$ and Question1 has the trivial answer $f=I_p$ and $C=0$. Question2 can be tackled through substitutions and integration by parts, assuming the continuity and unbounded-ness of $g(x)$. Commented Sep 27, 2019 at 8:37
• @Jack D'Aurizio I mentioned $C>0$ Commented Sep 27, 2019 at 8:56

For the first part, use the double angle formula to have $$\sin^2(x+1)=\frac 12 \left(1-\cos(2(x+1))\right)$$ Then use $$\cos(t)=\frac {e^{it}+e^{-it}}2$$ to face the exponential integral function.
You should end with something like $$I(a)=\int_0^\infty \frac{ \sin ^2(x+1)} {(x+1)^{a}}\,dx=\frac{1}{2 (a-1)}-\frac{1}{4} (E_a(2 i)+E_a(-2 i))$$ provided $$\Re(a)>1$$.