We yet get another case where π and e are directly connected. To find this integral we can do an ordinal variable substation: Now, since the variables are independent we can basically do the opposite we did above when we put the integrals inside each other. So to integrate over the whole surface instead of integration from minus to plus infinity over both x and y we will integrate for a radius from 0 to infinity and an angle from 0 to 2 π. We have thatįor our ΔA we will use ΔA=rΔθΔr as seen in the figure. Seen from above we basically integrate the whole area by adding small areas ΔA=ΔxΔy. Replacing t t with t2 t 2, inversing, and integrating from 0 0 to x x, gives a beautiful tan1 tan 1. Since the integral with respect to y can be seen as a constant seen from the respect to x, then we can move the whole integral inside the other. We have easily reached the conclusion that integral from 0 0 to x x of et2dt e t 2 d t has a limit somewhere between 0 0 and /2 / 2, as we used a little trick, precisely the inequality et > t + 1 e t > t + 1 for every real x x. So we start by making the integral into a two integrals, one in the x– and the y-direction. So what to do? The trick is to realize that an infinite rectangle is the same as an infinite square. The problem is that you cannot find a primitive function to the integrand, no matter what you do. This method is, I think, attributed to Gauss. This is definitely one of the classical difficult integrals. Next page : The integral of e^(-x^2) from –infinity to infinity a second way Previous page : The integral of ln(x+1)/(x^2+1) dx from 0 to 1
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