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This might be a duplicate but I couldn't find it. Anyhow I'm having trouble integrating $$\int^t_0 e^{x^2}dx$$ I found it in an introductory chapter of a differential equations book and perhaps it assumes I know an integration technique I haven't been introduced to(I only know integration techniques: by parts, u-sub., trig subst. partial fractions )? I have been asked to verify $e^{x^2} \cdot \int^t_0 e^{x^2}dx + c_1e^{x^2}$ as a solution to a differential???

I put it into an online calculator and got the result $e^{x^2}F(x)+c$ What function does F(x) stand for? The link for result is here:https://www.symbolab.com/solver/integral-calculator/%5Cint%20e%5E%7Bx%5E2%7Ddx/?origin=enterkey

I tried integration by parts and so far I got:

Let $u = e^{x^2}$, $du= 2xe^{x^2}dx$ and $dv=dx , v=x$

So substituting in $u \cdot v-\int vdu$ we get $$e^{x^2} \cdot x - \int x\cdot 2xe^{x^2}dx$$

Integrating by parts:$$\int x\cdot 2xe^{x^2}dx$$

I get $$e^{x^2} \cdot x^2 - \int 2x^3e^{x^2}dx$$

where $u_2=x^2, du_2=2xdx$ and $dv_2=2xdx, v_2=x^2$

Repeated integration by parts doesn't seem like a way to solving this. Does someone know of a way to integrate this?

Appreciate it.

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marked as duplicate by Simon S, user223391, Jack, user228113, Harish Chandra Rajpoot Dec 26 '15 at 4:51

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    $\begingroup$ This function cannot be integrated by elementary methods. However, there are some pretty ways to integrate it as a definite integral over certain bounds. $\endgroup$ – GiantTortoise1729 Dec 25 '15 at 23:27
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    $\begingroup$ This is a famous example of a function that does not have an elementary antiderivative. Here "elementary" means a function made up of polynomials, powers and roots, exponentials, logarithms, trigonometric functions and their inverses. $\endgroup$ – vonbrand Dec 25 '15 at 23:28
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    $\begingroup$ @Red: See WA for the imaginary error function result and also look below for Dawson Integral. $\endgroup$ – Moo Dec 25 '15 at 23:29
  • $\begingroup$ @GiantTortoise1729 What would some of those ways be? Actually I was the problem had it as x(upper limt) and 0 (lower limit). $\endgroup$ – Red Dec 25 '15 at 23:30
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    $\begingroup$ That is not as hard - you can differentiate them without a closed form, using FTC on the integral. $\endgroup$ – TokenToucan Dec 25 '15 at 23:46
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The integral in question can be written using a fairly common special function known as the "Error Function", which has the formula $$\operatorname{Erf(t)} = \frac{2}{\sqrt{\pi}} \int_0^t e^{-x^2} dx$$ We then note the derivative of $\frac{-i \sqrt{\pi}}{2} \operatorname{Erf(it)}$, which we get as $$-i \frac{d}{dt} \operatorname{Erf(it)} = \frac{-2i \sqrt{\pi}}{2 \sqrt{\pi}} e^{-(it)^2} \frac{d}{dt} (it) = e^{t^2}$$ Since $\operatorname{Erf(0)} = 0$, your definite integral becomes a simple antiderivative, which we get to be $\color{red}{\frac{-i \sqrt{\pi}}{2} \operatorname{Erf(it)}}$

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