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Wolfram Alpha isn't able to calculate this integral (I don't have mathematica, but I have Wolfram Pro).

$$\int_{0}^{a} \frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx \ \ \ , \ b>a$$

This is for a physics problem. I'd appreciate either a solution or the knowledge that the integral is non-soluble (which would indicate that I need to find some symmetry that I haven't seen yet). Thanks!

Simon S
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Zane Dufour
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    This is eliptic integral and can be solved by Euler's first and second substitutions. There are two cases, one when under square root quadratic polynomial has a real solution and other if it doesn't. Maybe that is the reason why Wolfram can't solve it. Try to put specific values for a and b. For more information go with this link: http://en.wikipedia.org/wiki/Euler_substitution – pointer Nov 10 '14 at 20:52
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    If you make suitable "linear" change of coordinate, you can turn the integral to the form $$\alpha \int_{\beta}^{\gamma} \frac{1}{\sqrt{1+u^2}} du = \alpha \left( \sinh^{-1}(\gamma) - \sinh^{-1}(\beta)\right)$$ where $\alpha, \beta, \gamma$ are constants. If you go through the algebra, you will get

    $$\frac{1}{\sqrt{2}}\left[\sinh^{-1}\left(\frac{b+a}{b-a}\right) - \sinh^{-1}(1)\right]$$

     – achille hui Nov 10 '14 at 20:56
  • Analytic solution existence for definite integrals are an open problem in mathematics; solving it would solve many open problems in transcendence theory. You can try Axiom to see if there is an indefinite integral, as it fully implements Risch.

    Also, it looks like wolfram alpha can solve the indefinite integral, so you can just plug your numbers in for the limits of integration. Not sure why it's giving up on the definite integral.

     – dezakin Nov 10 '14 at 21:04

3 Answers3

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I was able to specify the closed-form with Maple.

$$\int_{0}^{a} \frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx$$ equals to $$\frac {\sqrt 2}2 \left(\ln \left( \left( \sqrt {2}+2\,{\operatorname{csgn}} \left( a-b \right) \right) \left( a-b \right) \right) - \ln \left(2\,\sqrt {{a}^{2}+{b}^{2}}-\sqrt {2}\left(a+b\right) \right) \right), $$

where $\operatorname{csgn}$ is the complex signum function.

If we assume that $b>a$, then we could simplify it into the form

$$\frac {\sqrt 2}2 \left( \ln \left( b-a \right)+\ln \left( \sqrt {2}-1 \right) -\ln \left( \sqrt {2} \sqrt {{a}^{2}+{b}^{2}}-a-b \right) \right).$$

user153012
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Wolfram Alpha evaluates the indefinite integral as

$$\int \frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx =\frac{\log\left(\sqrt2 \sqrt{2x^2 - 2(a+b)x + a^2 + b^2} + 2x - a - b \right)}{\sqrt2}+\mbox{constant}.$$

You can get to Maple's closed-form solution from this if you plug in the limits of the integration, but personally I think the arccosh solution is neater.

David K
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WolframAlpha give this : http://www.wolframalpha.com/input/?i=integrate+dx%2Fsqrt%28%28x-a%29%5E2%2B%28x-b%29%5E2%29+ $$\int\frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx =\frac{1}{\sqrt 2} \ln\big(\sqrt{2a^2-4ax+2b^2-4bx+4x^2}-a-b+2x\big)+c$$ Then, it is easy to compute the defined integrtal : $$\int_0^a\frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx =\frac{1}{\sqrt 2} \ln \frac{\sqrt{2a^2+2b^2-4ab}+a-b}{\sqrt{2a^2+2b^2}-a-b}$$

JJacquelin
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