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Using double integrals to evaluate single integrals 
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#1
Feb313, 05:45 PM

P: 333

In most calculus textbooks, they use double integrals to evaluate the Gaussian integral. Where did they get the idea  or how did they choose the two variable function [tex]e^{(x^2+y^x)}[/tex] to evaluate it?
I guess this is related...but if you were given a fairly hairy integral and it was suggested that you write the integrand as an integral, how would you go about doing so? The example I have in mind is [tex]\int^2_0 arctan(\pi x)arctan(x) dx[/tex]. The only way I know how to do that is by integration by parts 


#2
Feb413, 01:54 PM

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#3
Feb413, 02:28 PM

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PF Gold
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If you have [itex]\int_a^b F(x)dx[/itex] and F'(x)= f(x) then, by the fundamental theorem of Calculus,[itex]F(x)= \int_p^x f(y)dy[/tex] for some p so [itex]\int_a^b F(x)dx= \int_a^b\int_p^x f(y)dydx[/itex]



#4
Feb1513, 08:53 AM

P: 333

Using double integrals to evaluate single integrals
I tried it on the integral I posted and I've probably misunderstood something, but it doesn't seem to simplify it at all (still have to use integration by parts). And secondly, how do you choose the lower bound p? What I got was: [tex]\int_0^{\pi} \int_0^2 arctan(\alpha x)arctan(x)dxd\alpha[/tex] I ended up choosing 0 as a lower bound cause that's when the integrand is 0.....hmmm...I think I've got down the wrong track here 


#5
Feb1513, 03:26 PM

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Your other example seems to be completely unrelated. 


#6
Feb1513, 03:48 PM

HW Helper
P: 1,391

$$\int_0^2 dx~(\mbox{arctan}(\pi x)  \mbox{arctan}(x) ) = \int_0^2 dx~\left. \mbox{arctan}(yx)\right^{y=\pi}_{y=1}.$$ You want to replace the ##\left. \mbox{arctan}(yx)\right^{y=\pi}_{y=1}## with an integral. Do you think you know what to do now? 


#7
Feb1613, 01:45 AM

P: 828

There are integrals that are difficult to evaluate, but sometimes the integrand can be modified by adding another variable. Then, one can use, say, Fubini's Theorem to evaluate the original integral by switching order of integration and comparing. So, perhaps the integral he posted can be computed that way. 


#8
Feb1613, 05:15 AM

P: 233

so if [tex]A=\int e^{x^2}dx[/tex] then [tex]A^2=(\int e^{x^2}dx)(\int e^{y^2}dy)=\iint e^{(x^2+y^2)}dxdy[/tex] so[tex] A= \sqrt{\iint e^{(x^2+y^2)}dxdy}[/tex] 


#9
Feb2413, 02:22 PM

P: 333

I'm not sure how you knew to replace the original integrand with arctan(yx) evaluated from y=1 to y=π 


#10
Feb2413, 11:05 PM

HW Helper
P: 1,391

$$ \int_a^b dy~\frac{d}{dy} f(yx)= \left.f(yx)\right_{y=a}^{y=b} = f(bx)  f(ax).$$ I wrote it as ##\left. \mbox{arctan}(yx)\right_{y=1}^{y=\pi}## rather than as an integral so that you could try and fill in the last step yourself. Does this make more sense? 


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