Answer
The statement is True:
The given improper integral
$$
\int_{1}^{\infty} \frac{1}{x^{\sqrt 2}} d x =\frac{1}{\sqrt {2}-1}
$$
is convergent.
Work Step by Step
The given integral
$$
\int_{1}^{\infty} \frac{1}{x^{\sqrt 2}} d x
$$
is an improper integral, hence
$$
\begin{split}
\int_{1}^{\infty} \frac{1}{x^{\sqrt 2}} d x & = \lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{x^{\sqrt 2}} d x \\
& =\lim _{t \rightarrow \infty} \left[ \frac{x^{-\sqrt {2}+1}}{-\sqrt {2}+1} \right]_{1}^{t}\\
& =\lim _{t \rightarrow \infty} \frac{1}{-\sqrt {2}+1} \left[ \frac{1}{t^{\sqrt {2}-1}}-1 \right] \\
& = \frac{1}{\sqrt {2}-1}
\end{split}
$$
Thus the given improper integral
$$
\int_{1}^{\infty} \frac{1}{x^{\sqrt 2}} d x =\frac{1}{\sqrt {2}-1}
$$
So, the integral is convergent.
