Nov 11, 2015 · The integral from $-\infty$ to $\infty$ is just twice this. So boom. If you want, you can rewrite $e^ {ix^2}=\cos (x^2)+i\sin (x^2)$ and equate the real and imaginary parts in the last equation …

May 23, 2024 · Definition of the Riemann Integral: I'm not sure how necessary this is, but I feel it might be worth going over nonetheless, just in case. The definition for the Riemann integral is actually not …

Oct 12, 2017 · This integral is one I can't solve. I have been trying to do it for the last two days, but can't get success. I can't do it by parts because the new integral thus formed will be even more difficult to …

Hence, for the integral you can just build this block matrix with B= I B = I, compute the matrix exponential of it, then extract the top right block. In a more "closed" form:

Feb 9, 2011 · Try using the iterative definition of the Cantor function, which gives a sequence of functions that converge uniformly to the Cantor function; then integrate each of those (or try a few …

If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect.

Nov 23, 2025 · Closed 14 days ago. The integral $$\int_0^\infty \left (\dfrac {\sin x} {x}\right)^3dx$$ can be computed by contour integration.

Feb 17, 2025 · The noun phrase "improper integral" written as $$ \int_a^\infty f (x) \, dx $$ is well defined. If the appropriate limit exists, we attach the property "convergent" to that expression and use …

Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.

The main result gives a necessary and sufficient condition under which the limit can be moved inside the integral. The exact condition is somewhat complicated, but it's strictly weaker than uniform integrability.