+33657 It depends on the limits of the integration. If the limits are ±infinity then the integral can be done analytically - see below. However, if at least one of the limits is finite then the integral can only be done numerically (an exception is if one of the limits is zero, then the integral from zero to infinity is just half of the integral from -infinity to infinity).
Here is a brief derivation of the integral of e-x^2 from - infinity to infinity:
LetI=\int_{-\infty}^{\infty}e^{-x^2}dxBecausexisadummyvariablewecanalsowritethisasI=\int_{-\infty}^{\infty}e^{-y^2}dyMultiplythetwotogethertogetI^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2}dxdyNowifwethinkofxandyasbeingaxesonastandardCartesiangraph,thentheright−handsideisjusttheintegraloverthewholeplaneof$e−(x2+y2)$.Changetopolarcoordinates,$r,θ$wherex = r.cos\theta, y = r.sin\theta.Thelimitsoftheintegrationbecome$0$to$∞$for$r$and$0$to$2π$for$θ$.$dxdy$becomes$r.drdθ$.So:
I2=∫2π0∫∞0e−r2r.drdθ\\
I2=2π∫∞0e−r2r.dr\\
I2=2π∫∞0e−r212dr2\\
I2=π∫∞0e−zdz\\
where I've substituted $z$ for $r^2$ in that last integral. So:
I2=π
as $\int_{0}^{\infty}e^{-z}dz=1$
Finally, $I=\sqrt\pi$ or
∫∞−∞e−x2dx=√π$$
+33657 It depends on the limits of the integration. If the limits are ±infinity then the integral can be done analytically - see below. However, if at least one of the limits is finite then the integral can only be done numerically (an exception is if one of the limits is zero, then the integral from zero to infinity is just half of the integral from -infinity to infinity).
Here is a brief derivation of the integral of e-x^2 from - infinity to infinity:
LetI=\int_{-\infty}^{\infty}e^{-x^2}dxBecausexisadummyvariablewecanalsowritethisasI=\int_{-\infty}^{\infty}e^{-y^2}dyMultiplythetwotogethertogetI^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2}dxdyNowifwethinkofxandyasbeingaxesonastandardCartesiangraph,thentheright−handsideisjusttheintegraloverthewholeplaneof$e−(x2+y2)$.Changetopolarcoordinates,$r,θ$wherex = r.cos\theta, y = r.sin\theta.Thelimitsoftheintegrationbecome$0$to$∞$for$r$and$0$to$2π$for$θ$.$dxdy$becomes$r.drdθ$.So:
I2=∫2π0∫∞0e−r2r.drdθ\\
I2=2π∫∞0e−r2r.dr\\
I2=2π∫∞0e−r212dr2\\
I2=π∫∞0e−zdz\\
where I've substituted $z$ for $r^2$ in that last integral. So:
I2=π
as $\int_{0}^{\infty}e^{-z}dz=1$
Finally, $I=\sqrt\pi$ or
∫∞−∞e−x2dx=√π$$
+118703 hi Alan,
I have had to split up LaTex before. It seems to happen when the output is long.
Today and I think yesterday the \frac was not working properly.
I'll try it now to show you
1-\frac{3}{(x+5)^2}
displays as - NOW IT WORKS! It wouldn't work when i was anwering Stu's question.
1−3(x+5)2
Have you managed to get begin{align} or begin{equarray} (I used to use this one all the time and now I am not even sure that the command is correct) to work? I have really missed not having those.
Thanks.
+33657 I haven't used either of begin{align} or begin{equarray}, I'm afraid.
I suspect that my problems are more likely to be of my own making in the main as I'm still very much a beginner with LaTeX. I find it easier to use Texmaker and then copy and paste into the LaTex Formula box than write it directly here.
+118703 I didn't think that would work properly - they are not usually that compatible.
Do copy in everything or just the main command parts? I'll have to try that.
Thanks.
To dispaly properly we really need some version of align! I'll leave him alone at the moment but when other problems lessen I will ask Andre about it. Maybe by then I (or you) will have worked it out. 
