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I need to solve the integral $\int^x \frac{1}{t^2}e^{t^2}dt. $ The answer should be $\sqrt{\pi}\left(\text{erfi}(x)\right)-\frac{e^{x^2}}{x} + C$

What does $\text{erfi}(x)$ mean? Can anybody explain how to solve this integral step-by-step? Thanks!

I need to solve the 2nd order differential equation $y'' -2xy'+2y=0 $ using that $y_1=Cx$. The solution of $y_2$ should then be \begin{align}y_2 &=y_1\int^x\frac{1}{{y_1(t)}^2}\exp \left(-\int^t{-2s}\,ds\right) \,dt\\ &=Cx\int^x\frac{1}{(Ct)^2}\exp \left(\int^t{(2s)}\,ds\right) \,dt\\ &=\frac{x}{C}\int^x\frac{1}{t^2} \exp \left(\int^t{(2s)}\,ds\right) \,dt\\ &= \frac{x}{C}\int^x\frac{1}{t^2}e^{t^2}\, dt \end{align}

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    tnx. My ultimate goal is to write down the solution of this differential equation as correct as possible :-)2012-12-02

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You are correct, $y_2=\frac{x}{C}\int\frac{e^{x^2}}{x^2}\, dx $. This indefinite integral can't be written in terms of elementary functions (polynomials, exponential, trigonometric functions and their inverses). It is useful however to express it in terms of power series. Indeed since $e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$ we easily see that \begin{equation}\int\frac{e^{x^2}}{x^2}\, dx=\int\frac{1}{x^2}\sum_{n=0}^{\infty}\frac{(x^2)^n}{n!}\, dx=\int\sum_{n=0}^{\infty}\frac{x^{2n-2}}{n!}\, dx=\sum_{n=0}^{\infty}\int\frac{x^{2n-2}}{n!}\, dx=\sum_{n=0}^{\infty}\frac{x^{2n-1}}{n!(2n-1)}\, dx \end{equation} Therefore, \begin{equation}y_2=\frac{x}{C}\int\frac{e^{x^2}}{x^2}\, dx=\frac{1}{C}\sum_{n=0}^{\infty}\frac{x^{2n}}{n!(2n-1)}\, dx \end{equation} In order to study such sums (and especially integrals with terms like $e^{x^2}$) that are very improtant for probability and statistics, we defined the error function $erf$ and the imaginary error function $erfi$. The Wikipedia entry for Error Function might be of interest to you. In the course of differential equations, you shall encounter other differential equations, the solutions of which may not elementary functions but can be expressed as a power series.

EDIT: We shall reduce this to the error function: \begin{gather}\int\frac{e^{x^2}}{x^2}\, dx=-\int\left(\frac{1}{x}\right)^{\prime}e^{x^2}\, dx=-\frac{e^{x^2}}{x}+\int\frac{1}{x}(e^{x^2})^{\prime}\, dx=-\frac{e^{x^2}}{x}+\int\frac{1}{x}e^{x^2}2x\, dx\Rightarrow \\\int\frac{e^{x^2}}{x^2}\, dx=-\frac{e^{x^2}}{x}+2\int e^{x^2}\, dx\end{gather} I think the OP can handle the rest

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    @Hempo Let me update my answer.2012-12-02