Skip to main content

calculus and analysis - Bad performance of Integrate (and WolframAlpha) for an Integral of Bessel function of the first kind: Version 11 edit


Version 11 Edit


The issue still remains:


Integrate[BesselJ[0, x], {x, 0, ∞}] // Timing
(* {29.8125, 1} *)

$Version


(* "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" *)


Original post


The following returns unevaluated in WolframAlpha. Also in my machine Mathematica needs quite a lot of time to compute it.


In[720]:= Integrate[BesselJ[0, x], {x, 0, \[Infinity]}] // Timing
Out[720]= {26.520170, 1}

I am puzzled because I thought such integrals are well tabulated in Mathematica.


At least the numerical integration is performed very quickly.


In[721]:= NIntegrate[BesselJ[0, x], {x, 0, \[Infinity]}] // Timing
Out[721]= {0.826805, 1.}


Any ideas about this performance of Mathematica and WolframAlpha?


EDIT


As J. M. is back suggested the following gives the result almost immediately


In[734]:= LaplaceTransform[BesselJ[0, t], t, s] /. s -> 0 // Timing
Out[734]= {0., 1}

For me the poor behavior of Integrate here still remains. What is even worse are the differences between the result of Mathematica and WolframAlpha.


In[761]:= LaplaceTransform[BesselJ[0, t], t, s]
Out[761]= 1/Sqrt[1 + s^2]


whereas WolframAlpha returns


1/(sqrt(1+1/s^2) s)

So you have to type


Limit[1/(sqrt(1+1/s^2) s),s->0]

in order to get finally 1.


EDIT 2


In[2]:= Quit[]


In[1]:= Integrate[BesselJ[0, x], {x, 0, \[Infinity]},
GenerateConditions -> False] // Timing
Out[1]= {26.301769, 1}

In[2]:= Quit[]

In[1]:= Integrate[BesselJ[n, x], {x, 0, \[Infinity]},
GenerateConditions -> False] // Timing
Out[1]= {0.109201, 1}


In[2]:= Quit[]

In[1]:= Integrate[BesselJ[n, x], {x, 0, \[Infinity]},
GenerateConditions -> True] // Timing
Out[1]= {8.049652, ConditionalExpression[1, Re[n] > -1]}

Compare the times 8.049652 (with n and GenerateConditions->True !) and 26.301769 (with n=0).


Rather funny:-)!


EDIT 3 Even more weird...



In[2]:= Quit[]

In[1]:= Integrate[BesselJ[n, x], {x, 0, \[Infinity]},
Assumptions -> n > 0] // Timing
Out[1]= {22.417344, 1}

Compare the times of Assumptions and GenerateConditions->True.


EDIT 4


Apparently something goes very wrong. With a clear kernel I got


Table[Timing[Integrate[BesselJ[0, x], {x, 0, Infinity}]], {i, 1, 10}]

Table[Timing[Integrate[BesselJ[1, x], {x, 0, Infinity}]], {i, 1, 10}]
Table[Timing[Integrate[BesselJ[2, x], {x, 0, Infinity}]], {i, 1, 10}]
Table[Timing[Integrate[BesselJ[3, x], {x, 0, Infinity}]], {i, 1, 10}]

Results


For m >= 4 Mathematica gives the 1 for m odd (same behavior as for m=3). For m even also gives 1 in all cases but needs more than 20 sec per average to evaluate the integral.


I cannot find a reason to explain this behavior which other more experienced users have called it buggy. I think Mathematica makes use of this reference. So apparently something is treated differently for n even.


EDIT 5 I_Mariusz pointed out a workaround. I added here in order to show its performance (if any) in Timing (and since on one hand our comments had misprints and on the other I did not know that Integrate can be compiled).


With a clear kernel I got:


int = Compile[{{x, _Real}}, 

Integrate[BesselJ[0, x], {x, 0, Infinity},
GenerateConditions -> False]];

Table[int[x] // Quiet // Timing, {10}]


{{26.254968, 1}, {24.897760, 1}, {24.897760, 1}, {20.046128, 1}, {0.046800, 1}, {0.046800, 1}, {0.062400, 1}, {0.062400, 1}, {0.062400, 1}, {0.062400, 1}}



At least now we get 1's in the whole loop!



Answer




I think this is not a answer but can serve as a summary of the useful workarounds. Still we do not have a rigorous explanation of the buggy (?) behavior.


Since


Integrate[BesselJ[0, x], {x, 0, \[Infinity]}] // Timing
(*{27.112974, 1}*)

J. M. is back suggested the following which gives the result almost immediately.


LaplaceTransform[BesselJ[0, t], t, s] /. s -> 0 // Timing
(*{0., 1}*)

In WolframAlpha we will get 1/(sqrt(1+1/s^2) s) from the LaplaceTransform so we need in addition Limit[1/(sqrt(1+1/s^2) s),s->0].



Integration with generic n gives the correct result but still the time is very large.


Integrate[BesselJ[n, x], {x, 0, \[Infinity]}] // Timing
(*{11.731275, ConditionalExpression[1, Re[n] > -1]}*)

Integrate[BesselJ[n, x], {x, 0, \[Infinity]},
Assumptions -> n \[Element] Reals] // Timing
(*{29.780591, ConditionalExpression[1, n > -1]}*)

As a workaround we can ignore the conditions


ClearSystemCache[]

Integrate[BesselJ[n, x], {x, 0, \[Infinity]},
GenerateConditions -> False] // Timing
(*{0.093601, 1}*)

or more rigorous (after B. Hanlon)


 Simplify[Integrate[BesselJ[n, x], {x, 0, \[Infinity]}, GenerateConditions -> True], n > -1] // Timing
(*{11.778075, 1}*)

Another interesting workaround is by I_Mariusz


int = Compile[{{x, _Real}}, 

Integrate[BesselJ[0, x], {x, 0, Infinity},
GenerateConditions -> False]];

Table[int[x] // Quiet // Timing, {10}]


{{26.254968, 1}, {24.897760, 1}, {24.897760, 1}, {20.046128, 1}, {0.046800, 1}, {0.046800, 1}, {0.062400, 1}, {0.062400, 1}, {0.062400,1}, {0.062400, 1}}



Notice that


Table[Timing[Integrate[BesselJ[0, x], {x, 0, Infinity}]], {i, 1, 10}]


returns only the first three iterations corrected.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...