Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Lists of integrals
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Integrals with a singularity=== When there is a [[Singularity (mathematics)|singularity]] in the function being integrated such that the antiderivative becomes undefined at some point (the singularity), then ''C'' does not need to be the same on both sides of the singularity. The forms below normally assume the [[Cauchy principal value]] around a singularity in the value of ''C'', but this is not necessary in general. For instance, in <math display="block">\int {1 \over x}\,dx = \ln \left|x \right| + C</math> there is a singularity at 0 and the [[antiderivative]] becomes infinite there. If the integral above were to be used to compute a definite integral between β1 and 1, one would get the wrong answer 0. This however is the Cauchy principal value of the integral around the singularity. If the integration is done in the complex plane the result depends on the path around the origin, in this case the singularity contributes β''i''{{pi}} when using a path above the origin and ''i''{{pi}} for a path below the origin. A function on the real line could use a completely different value of ''C'' on either side of the origin as in:<ref>[[Serge Lang]] . ''A First Course in Calculus'', 5th edition, p. 290</ref> <math display="block"> \int {1 \over x}\,dx = \ln|x| + \begin{cases} A & \text{if }x>0; \\ B & \text{if }x < 0. \end{cases} </math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)