Is this integral solvable?

I am trying to solve the below mentioned integral. Is it possible to solve this with mathematica. If not, can I get an approximation for this by introducting the variable not to take certain values ?

Integrate[
4 x/π^2 ArcCos[(y^2 + 3)/(4 y)] ArcCos[(x^2 + y^2 -
1)/(2 x y)], {x, 0, 1}, {y, 1, 2}]

• Some spaces would do a lot to clarify your expression. Even better would be Mathematica code - have you tried anything yet? – Yves Klett Sep 22 '14 at 7:59
• I have edited the post with a newer image. I do not have the software installed yet but I am looking to find out whether this solvable or not with mathmetica. I want to have value/solution to put in an expression – Waqas Sep 22 '14 at 8:03
• Better than an image would be TeX code. That would at least eliminate some work for us. – Yves Klett Sep 22 '14 at 8:05
• I am sorry for not providing the code since I am in the process of installing and getting to use mathemtica only solve this integral problem. If I can have an information about the solvability, it would be great. I tried the online integral calculator but it says it does not have a solution. – Waqas Sep 22 '14 at 8:09
• You can try the free Wolfram cloud, it has Mathematica there for free wolfram.com/programming-cloud/pricing click on the free option. Also can try Wolfram alpha, it is supposed to be able to do integration as well. – Nasser Sep 22 '14 at 8:13

Judging from the shape,

Plot3D[4 (x y)/\[Pi]^2 ArcCos[(y^2 + 3)/(4 y)] ArcCos[(x^2 + y^2 -
1)/(2 x y)], {x, 0, 1}, {y, 1, 2}]


and from

Reduce[
{
4 (x y)/\[Pi]^2 ArcCos[(y^2 + 3)/(4 y)] ArcCos[(x^2 + y^2 - 1)/(2 x y)] == 0
, 0 <= x <= 1
, 1 <= y <= 2
}, {x, y}
]


0 < x <= 1 && (y == 1 || y == 1 + x)

the expression should be integrable in the range {x, 0, 1}, {y, 1, 1 + x}

NIntegrate[
4 (x y)/\[Pi]^2 ArcCos[(y^2 + 3)/(4 y)] ArcCos[(x^2 + y^2 -
1)/(2 x y)], {x, 0, 1}, {y, 1, 1 + x}, WorkingPrecision -> 20]


0.053240519638085381131

• I think you're missing a y in the inner integral. – Netsie Sep 22 '14 at 9:31
• @Netsie, you are correct, I have corrected that now. – rhermans Sep 22 '14 at 9:32
• Thank you so much for the solution. I have actually observed that the actual integral equation has always the 1+x upper range for y instead of the value which I had earlier shared. I want to ask if the upper limit of the y integral is x+1, then this integral is integrable over the entire range ? – Waqas Sep 23 '14 at 0:27
• @Waqas, The expression is real and well behaved in {x, 0, 1}, {y, 1, 1 + x} , and complex for y>1+x. Does this answer your question? – rhermans Sep 23 '14 at 6:12
• Yes it surely does and I am pretty relieved.As I am not that efficient in mathematical language, can you elaborate a bit more what does well behaved mean ? – Waqas Sep 23 '14 at 10:56