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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 $[1,\infty)$

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    $\begingroup$ @JackD'Aurizio: why doesn't $I(x;a)$ depend on $x$? Isn't it the antiderivative of a function of $x$? $\endgroup$
    – YiFan Tey
    Commented Sep 27, 2019 at 8:28
  • $\begingroup$ Someone downvoted ? Is something wrong ? $\endgroup$
    – bambi
    Commented Sep 27, 2019 at 8:37
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    $\begingroup$ 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 $g(x)$. $\endgroup$ Commented Sep 27, 2019 at 8:37
  • $\begingroup$ @Jack D'Aurizio I mentioned $C>0$ $\endgroup$
    – bambi
    Commented Sep 27, 2019 at 8:56

1 Answer 1

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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 $\Re(a)>1$.

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  • $\begingroup$ @ Claude Leibovici thank you for the answer . $\endgroup$
    – bambi
    Commented Sep 27, 2019 at 8:35
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    $\begingroup$ @Bambi If you think it's a satisfactory answer, you can upvote it and mark it as accepted (by clicking the tick mark below the vote buttons). $\endgroup$
    – YiFan Tey
    Commented Sep 27, 2019 at 9:26
  • $\begingroup$ @YiFan I tried to upvote but not enough reputation ( so only recorded ,not shown) and I'm looking for any further comment on question 2 as it's my main concern . $\endgroup$
    – bambi
    Commented Sep 27, 2019 at 9:30
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    $\begingroup$ @Bambi Sure, just making sure you knew the mechanics of the site :) Welcome to MSE! $\endgroup$
    – YiFan Tey
    Commented Sep 27, 2019 at 9:31
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    $\begingroup$ @YiFan thank you . (And also thank you for supporting earlier ) $\endgroup$
    – bambi
    Commented Sep 27, 2019 at 9:33

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