While working on a differential equation I stumbled on this integral: $$\int_0^x\frac{e^{-t^2}}{a^2+t^2}dt,$$ where $a\in \mathbb{R_{>0}}$.

At first glance it looks so simple that it's ridiculous that I can't find closed form expression for it.

Can someone help me with this beast?

$Edit:$From the comments, it doesn't seem to exist a closed form. I tried a series expansion by writing the integrand as $e^{-t^2}g(t)$ and take the Taylor expansion of the function $g(t)$. It's a good approximation and it's possible to integrate the expression found but it's only valid for small values of $x$ since the series diverge.

  • $\begingroup$ I'm not sure it has a closed expression. Try to exchange the integrals, hopefully the other direction gets you some $t$ as multiplier. $\endgroup$ – Berci Jun 13 '13 at 23:25
  • $\begingroup$ Or, perhaps, do you want its limit as $x\to\infty$? $\endgroup$ – Berci Jun 13 '13 at 23:27
  • $\begingroup$ @Berci Well, there's no other integration to be made. I have a ODE whose solution is $f(x)=\int_0^x \frac{e^{-t^2}}{a^2+t^2}dt+...$. So I can't take the limit. I'm sorry if I misled you with the way I wrote the question. $\endgroup$ – PML Jun 13 '13 at 23:32
  • $\begingroup$ I would guess that this does not have an expression in terms of elementary functions. $\endgroup$ – parsiad Jun 13 '13 at 23:33
  • 1
    $\begingroup$ @par I wasn't implying that it does. I was just answering the first two comments. $\endgroup$ – PML Jun 13 '13 at 23:36


Consider $\int\dfrac{t^{2n}}{a^2+t^2}dt$ ,

Let $t=a\tan\theta$ ,

Then $dt=a\sec^2\theta~d\theta$



$=\int a^{2n-1}\tan^{2n}\theta~d\theta$



Hence $\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{n!(a^2+t^2)}dt$



  • $\begingroup$ This is outstanding!Thank you very much, really. I hope you get many upvotes for the work you've put into this answer. Thank you once again. $\endgroup$ – PML Jun 15 '13 at 0:50

I'll tackle the case of the large values of $x$. Let's express the integral as follows:



The next step is evaluate the last integral by parts:



In the last expression, the first term dominates for large $x$ and we have approximately:

$$\int_0^x\frac{e^{-t^2}}{a^2+t^2}dt\approx I(a)-\frac{e^{-x^2}}{2x(a^2+x^2)}$$

We'll get a better result if we continue the integration by parts in the penultimate expression.

Now, consider


Let's introduce the parameter $b$ so that


and $I(a)=J(1)$

By differentiating with respect to $b$ we will see that $J$ satisfies the following differential equation:


Taking into account that

$$J(0)=\int_0^{\infty}\frac{dt}{a^2+t^2}=\frac{\pi}{2a}$$ we get a solution of this diff-eq:



$$I(a)=J(1)=\sqrt{\pi}\frac{e^{a^2}}{a}\int_{a}^{\infty}e^{-y^2}dy=\frac{\pi}{2}\frac{e^{a^2}}{a}\text{erfc}(a)$$ where $\text{erfc}(x)=\frac{2}{\sqrt{\pi}}\int_{x}^{\infty}e^{-y^2}dy$ is so called "the complementary error function"

  • $\begingroup$ I'm learning a lot with your answers. I thank you very much with the time you took on your answer. Your approach is very interesting and pedagogical. Thank you once again $\endgroup$ – PML Jun 17 '13 at 21:12

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