How Do I Integrate eSign in Jitterbit
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How To Integrate eSign in Jitterbit
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Templatize frequently used documents to save time and reduce the risk of common errors when sending out copies for signing.
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FAQs
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How do I calculate indefinite integral of [math]e^{x^2}[/math]?
Integration of e^(x^2)
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How do you use LaTeX mathematical notation for formulae on Quora?
From our Help Center article, How do you use LaTeX mathematical notation for formulae on Quora?:You can write LaTeX and have it styled correctly by using the Math formatting option in the Quora editor (see screenshot) or by using the keyboard shortcut.For more information about Quora’s features and frequently asked questions, check out our Help Center.
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What is the antiderivative of [math]e^{-x^2}[/math]?
The integral in this question can be expressed in terms of what is called a special function.The series expansion of [math]\displaystyle e^{-x^2}[/math] around [math]x = 0[/math] is expressed as (verified with Mathematica):[math]\displaystyle 1-x^2 +\frac{x^4}{2}-\frac{x^6}{6} +\frac{x^8}{24}-\frac{x^{10}}{120}+ …[/math]The series expansion above is symbolically written as:[math]\displaystyle e^{-x^2}=\sum _{n=0}^{\infty } \frac{\left(-x^2\right)^n}{n!}=\sum _{n=0}^{\infty } \frac{(-1)^n \left(x^2\right)^n}{n!}[/math]Using the last result above and integrating, we get:[math]\displaystyle \int e^{-x^2} \, dx = \int \left(\sum _{n=0}^{\infty } \frac{(-1)^n \left(x^2\right)^n}{n!}\right) \, dx=\sum _{n=0}^{\infty } \frac{(-1)^n x^{2 n+1}}{(2 n+1) n!}+C[/math]If we examine the infinite sum obtained above, and look at the series expansion of the error function which is given by:[math]\displaystyle \operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{n!(2n+1)}} \\ \displaystyle \qquad ={\frac {2}{\sqrt {\pi }}}\left(x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{10}}-{\frac {x^{7}}{42}}+{\frac {x^{9}}{216}}-\cdots \right),[/math]we see that the integral in this question can be expressed as:[math]\displaystyle \boxed{\int e^{-x^2} \, dx =\sum _{n=0}^{\infty } \frac{(-1)^n x^{2 n+1}}{(2 n+1) n!}+C = \frac{1}{2} \sqrt{\pi }* \text{erf}(x)+C}[/math]Below is a plot of the integration solution:Note that the solution to the given indefinite integral could be found by using the definition of the error function:[math]\displaystyle \text{erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-x^{2}}\,dx={\frac {2}{\sqrt {\pi }}}\int e^{-x^{2}}\,dx,[/math]so that we have:[math]\displaystyle \int e^{-x^{2}}\,dx = \frac {\sqrt{\pi}}{2} \text{erf}(x)+ constant[/math]Considering the representation of the error function via other general functions, the given integral can be also expressed in terms of the generalized hypergeometric function [math]\, _pF_q (a;b;x)[/math] in the following form:[math]\displaystyle \int e^{-x^2} \, dx= x* \, _1F_1\left(\frac{1}{2};\frac{3}{2};-x^2\right) + constant [/math]
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What is the integration of [math]e^{\sin x}[/math] ?
Thank You
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How do I integrate [math]\dfrac1{e^{2x} +e^{x}} ?[/math]
[math]I=\displaystyle \int \dfrac 1 {e^{2x}+e^x}dx[/math][math]=\displaystyle \int \dfrac 1 {e^x(e^x+1)}dx[/math][math]=\displaystyle \int \dfrac {(e^x+1)-e^x} {e^x(e^x+1)}dx[/math][math]=\displaystyle \int \dfrac {1} {e^x}dx-\int \dfrac {1} {e^x+1}dx[/math][math]=\displaystyle \int e^{-x}dx-\int \dfrac {e^{-x}} {1+e^{-x}}dx[/math][math]=\boxed{-e^{-x}+\ln (1+e^{-x})+C}[/math]
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Can I use Kik Messenger online without downloading the app?
Hi, I won’t type long article here. But here is how to use Kik login online no download. Kik Login Online Kik Login Online with No Download simply refers to using Kik on a computer without downloading an Android emulator. And Kik itself supports this features, but before you proceed, you have to create a new account but if you have an already registered account, it will be simple to login and begin Kik-ing. • Boot ON your PC / Smartphone • Now go to Kik login online website [ https://www.kik.com/login ] • Enter your username • Enter your Password • Tick ‘Remember Me’ • Click on Login That is it, you have finally log into your Kik account without any need of Android emulator.
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How do I integrate e^(e^x)?
Letif y = integral e^x thendy/dx = e^xSupposey = f(x) = e^xSo derivative of e^x is e^x itself. or in other words integral of e^x is e^x itself
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Frequently asked questions
How do i add an electronic signature to a word document?
When a client enters information (such as a password) into the online form on , the information is encrypted so the client cannot see it. An authorized representative for the client, called a "Doe Representative," must enter the information into the "Signature" field to complete the signature.
How do you know an electronic signature is real?
That you have the signature of an actual person that signed it.
And, of course, I do. Because that's the thing about an electronic signature. If you can't prove with something else that you were the actual person who actually signed it or that your physical signature is there, it becomes a fraud. That is, unless you could get a court to sign off on it, where the court would basically rule this electronic signature is a real signature, even if the electronic signature looks real to you. You can't be sure.
It's like the difference between a hologram and a hologram. It doesn't matter who put it there. They don't have to show a real hand to make it work. So, if you sign an electronic signature, if that person can't provide proof that the signature is theirs, it becomes fake. It becomes fraud.
So, in the case of electronic signatures, and there's an entire case that's been pending in the court for about ten years about, what to do about them, the judge actually said there was enough evidence in that case, which is sort of an interesting precedent for a lot of these kinds of cases. If you can show a court that an electronic signature can be faked, you could get a court ruling to allow you to make a copy of that signature and prove that the signature is fake. So that would solve that particular problem. It's not a complete fix by any stretch of the imagination, but it would solve that particular issue. So that would really solve one of the two problems, because then you could us...
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