I did notice that the integral of (1/sqrt(a

^{2}+u

^{2}) = arsinh (u/a) + C This is the closest form that I have found for a simple integral relevant to your problem.

Also, the integral of(1/(u*sqrt(a

^{2}+u

^{2})) = arcsch |(u/a)| + C where I think arcsch is an arc-hyperbolic cosecant.

All of which seems to be in sync with teliot's comment about Wolfram's advice to look at hyperbolic functions. However, hyperbolas have different mathematical properties than the curve you are studying and I'm not sure if they will reduce to your curve by, say, setting the distance between the two foci to 0.

Certainly not an "easy math problem."

Quote:gordonm888Recognizing that this was either an unsolvable integral (i.e., no analytic form for the anti-derivative) or at least one for which the solution is not widely known, I was just typing in notes in my posts as I studied it. I realized that my posts were not directly helpful, sho88, I was just trying to come up with a good idea on how to start the analysis.

I did notice that the integral of (1/sqrt(a^{2}+u^{2}) = arsinh (u/a) + C This is the closest form that I have found for a simple integral relevant to your problem.

Also, the integral of(1/(u*sqrt(a^{2}+u^{2})) = arcsch |(u/a)| + C where I think arcsch is an arc-hyperbolic cosecant.

All of which seems to be in sync with teliot's comment about Wolfram's advice to look at hyperbolic functions. However, hyperbolas have different mathematical properties than the curve you are studying and I'm not sure if they will reduce to your curve by, say, setting the distance between the two foci to 0.

Certainly not an "easy math problem."

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Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?

For a numerical approximation, this is exactly what I would do.Quote:ssho88

Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?

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Quote:gordonm888Nice. But with y = x^4, is there any convergence issue when x<1?

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Convergence issue is my concern too. Thanks for your comment.

Quote:teliotFor a numerical approximation, this is exactly what I would do.Quote:ssho88

Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?

link to original post

link to original post

Thanks for your comment.

However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?

Quote:ssho88Here is the integral solution:-

However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?

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Just the same comment I made twice before.

https://www.wolframalpha.com/input?i=arc+length+1%2FX&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22a%22%7D+-%3E%221%22&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22b%22%7D+-%3E%223%22