How Can I Integrate eSign in 1Password

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Integration of e^(x^2)

∫e^x/x dxIntegrate by parts, where u=1/x, and v′=e^x. Then u′=−1/x2, and v=e^x. So,∫e^x/x dx=e^x/x+∫e^x/(x^2)dxIntegrate by parts again, u=1/x^2, v′=e^x, so that u′=−2/x^3 and v=e^x. So,∫e^x/x dx=e^x/x+e^x/x^2+2∫e^x/x^3dxRepeat this process ad infinitum to get,∫e^x/x dx=e^x/x+e^x/x^2+2(e^x/x^3+3(e^x/x^4+4(e^x/x^5+⋯)))Expanding this gives,∫e^x/x dx=e^x/x+e^x/x^2+2e^x/x^3+6e^x/x^4+24e^x/x^5+⋯And factoring that gives,∫e^x/x dx=e^x/x(1+1/x+2/x^2+6/x^3+24/x^4+⋯)Now, considering the series itself, the ratio between the nth term and the (n−1)th term = nx. Eventually, n will be larger than x, so the ratio between successive terms will be positive, so (assuming x is positive), the series diverges, meaning that:∫e^x/xdx=e^x.x(∞)=∞

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Putting a little meat on the bones sketched out above, the answer is indeed "no."  It's often tough to get a precise answer though.  People will talk about it having no "closed form" solution, for example.  But the symbol [math]\int x^x dx[/math] is, itself, kind of a closed form.  So what does "closed form" really mean?Long story short, it means you can't write it down using a finite number of multiplications, additions, subtractions, divisions, roots, exponents, trigonometric functions, or logarithms.Long story a little longer, the answer involves a subject known as "differential Galois theory."  In a bit of a digression, Galois theory -- the plain, non-differential kind -- involves (among other things) studying roots of polynomials.  It's thanks to Galois theory that we know the quintic polynomial equation can't be solved by radicals, for example.What does that mean?  It means that you can't use the operations of addition, subtraction, multiplication, division, or roots in any (finite) combination to write down the solution to a generic fifth degree polynomial.  In fact, it's even worse... the solutions to some specific polynomials can't be written down that way, with or without a general formula.Given the similarity of the 5th degree polynomial result with the result you're asking about, it's not surprising that the two should be related.I'm hesitant to say much more, because Galois theory is kind of a long story.  It's not necessarily true that it's hard, but there's a ton of machinery that you have to develop, and Quora isn't the place to do it.  FWIW, this is the stuff that math majors study.  (And, to my knowledge, nobody else really does.)  So if this really gets you excited, consider being a math major.

Although [math]\frac{e^x}{x}[/math] is integrable, except at [math]x = 0[/math], the integral cannot be expressed in terms of elementary functions, so there isn’t a way to produce a conventional indefinite integral. Obviously, it is sometimes convenient to represent the solutions to such integrals in some compact way, and there are several special functions which can be used to do this:[math]\DeclareMathOperator{\Ei}{Ei} \DeclareMathOperator{\li}{li} \displaystyle \int \frac{e^x}{x} \,\mathrm dx = \Ei(x) = \li(e^x) = - \Gamma(0, -x)[/math]where [math]\Ei(x)[/math] is the exponential integral function, [math]\li(x)[/math] is the logarithmic integral function, and [math]\Gamma(s, x)[/math] is the incomplete gamma function. However, if you check the definitions of those functions, you will see that expressing the integral in these ways hasn’t actually done anything useful: those functions are defined as being the result of certain integrals, and so are simply handy means to notate functional solutions that we cannot write as expressions in elementary functions!Because we can’t write the integral in elementary functions, the only way to evaluate a definite integral of [math]\frac{e^x}{x}[/math] is to use numerical methods.