<s>[INST] Domain of definition of :$f(x)=\int_{0}^{\infty}\frac{t^x}{1+t^4 \sin ^2 t}dt$ let $f$ be a function of the variable $x$ defined as :  $$f(x)=\int_{0}^{\infty}\frac{t^x}{1+t^4 \sin ^2 t}dt$$  , Really i didn't succeed to determine the domain of definition of that function since i'm not able to get when does that converges , Then my question here is :What is the domain of definition of :$f(x)=\int_{0}^{\infty}\frac{t^x}{1+t^4 \sin ^2 t}dt$ ?
 [/INST] Since the integrand is positive, we can write
\begin{align}
f(x) &= \int \limits_0^\infty \frac{t^x}{1+t^4 \sin^2 (t)} \, \mathrm{d} t \\
&= \int \limits_0^{\pi/4} \frac{t^x}{1+t^4 \sin^2 (t)} \, \mathrm{d} t + \sum \limits_{k=1}^\infty \int \limits_{\left(k-\frac{3}{4}\right)\pi}^{\left(k-\frac{1}{4}\right)\pi} \frac{t^x}{1+t^4 \sin^2 (t)} \, \mathrm{d} t + \sum \limits_{k=1}^\infty \int \limits_{\left(k-\frac{1}{4}\right)\pi}^{\left(k+\frac{1}{4}\right)\pi} \frac{t^x}{1+t^4 \sin^2 (t)} \, \mathrm{d} t \\
&\equiv f_1 (x) + f_2 (x) + f_3 (x) \,
\end{align}
and estimate the three terms separately.
We have
$$ \frac{1}{1+2^{-9} \pi^4} \int \limits_0^{\pi/4} t^x \, \mathrm{d} t \leq f_1 (x) \leq \int \limits_0^{\pi/4} t^x \, \mathrm{d} t \, , $$
so $f_1(x) < \infty$ holds if and only if $x > -1$ .
For $k \in \mathbb{N}$ and $ \left(k-\frac{3}{4}\right)\pi \leq t \leq \left(k-\frac{1}{4}\right)\pi$ we have $\frac{1}{2} \leq \sin^2 (t) \leq 1$ , so
$$ \sum \limits_{k=1}^\infty \int \limits_{\left(k-\frac{3}{4}\right)\pi}^{\left(k-\frac{1}{4}\right)\pi} \frac{t^x}{1+t^4} \, \mathrm{d} t \leq f_2(x) \leq \sum \limits_{k=1}^\infty \int \limits_{\left(k-\frac{3}{4}\right)\pi}^{\left(k-\frac{1}{4}\right)\pi} \frac{t^x}{1+ \frac{1}{2} t^4} \, \mathrm{d} t \, . $$
We conclude that $f_2(x)$ is finite if and only if $x<3$ .
For the last term we can use
\begin{align}
f_3 (x) &\leq \pi^x \sum \limits_{k=1}^\infty \left(k+\frac{1}{4}\right)^x \int \limits_{\left(k-\frac{1}{4}\right)\pi}^{\left(k+\frac{1}{4}\right)\pi} \frac{1}{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4 \sin^2 (t)} \, \mathrm{d} t \\
&= \pi^x \sum \limits_{k=1}^\infty \left(k+\frac{1}{4}\right)^x \left[\frac{\arctan\left\{\sqrt{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4} \tan(x) \right\}}{\sqrt{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4}} \right]_{x= \left(k-\frac{1}{4}\right) \pi}^{x= \left(k+\frac{1}{4}\right) \pi} \\
&= 2 \pi^x \sum \limits_{k=1}^\infty \frac{\left(k+\frac{1}{4}\right)^x}{\sqrt{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4}} \arctan\left\{\sqrt{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4}\right\} \\
&\leq \pi^{x-1} \sum \limits_{k=1}^\infty \frac{\left(k+\frac{1}{4}\right)^x}{\left(k - \frac{1}{4}\right)^2}
\end{align}
and a similar estimate from below to find
$$ \frac{\pi^{x-1}}{2\sqrt{2}} \sum \limits_{k=1}^\infty \frac{\left(k-\frac{1}{4}\right)^x}{\left(k + \frac{1}{4}\right)^2} \leq f_3(x) \leq \pi^{x-1} \sum \limits_{k=1}^\infty \frac{\left(k+\frac{1}{4}\right)^x}{\left(k - \frac{1}{4}\right)^2} \, . $$
This implies $f_3(x) < \infty ~ \Leftrightarrow ~ x < 1$ .
Combining these results we see that $f(x) < \infty$ holds if and only if $-1 < x < 1$ , so $f$ is well-defined on the interval $(-1,1)$ .
</s>