I am evaluating an integral with constants that are not specified, but I am not sure why it takes so long for it to give an output, so I just decided to cancel the running. The integral is given by,
$\int_0^1 dy \frac{ c a^3 y^2 }{ ((1 - b^3 y^3)(1 - c^2 a^4 y^4))^{1/2} }$
d = 2;
z = 10;
b = a/z;
SumConvergence[(c a^(d + 1)
y^d)/((1 - b^(d + 1) y^(d + 1)) (1 - c^2 (a y)^(2 d)))^(1/2), y]
Integrate[(c a^(d + 1) y^
d)/((1 - b^(d + 1) y^(d + 1)) (1 - c^2 (a y)^(2 d)))^(1/2), {y, 0,
1}, Assumptions -> {c > 0, a > 0}]
d indicates dimensions so in this case I set for example, d=2, while a,b are constants (leave it open so I can put values later). In the end, I want to get an expression for "c" in terms of "a" (since "a" also gives "b") through evaluation of the integral.
UPDATE: I changed the integral expression a bit compared to my first post, now I tried doing the SumConvergence command and it returns a TRUE value so this new integral that I posted converges but I do not know why it does not return the condition of convergence. Also, the Integrate command still does not return anything even though the function converges.
ay
which should have a space between the two letters:a y
. Also, you don't need to statey>0
. But after fixing those errors, Mathematica states that the integral does not converge on {0,1}. But it does converge if you restricta<0
. $\endgroup$d = 2; z = 1; b = a/z; Integrate[(y^d (1 - (b y)^(d + 1))^1/2)/(1 - c^2 (a*y)^(2 d))^1/2, {y, 0, 1}, Assumptions -> c > 0 && a > 0]
produces $$\text{ConditionalExpression}\left[\frac{2 a^2 c+\log \left(\frac{2}{a^2 c+1}-1\right)-2 c^{3/2} \tan ^{-1}\left(a \sqrt{c}\right)+2 c^{3/2} \tanh ^{-1}\left(a \sqrt{c}\right)}{16 a^3 c^3},c<\frac{1}{a^2}\right] .$$ $\endgroup$z^1/2
and then tryz^(1/2)
orSqrt[z]
and see the difference. Introducing decimal points is probably a very bad idea. Once you fix your exponent problem you can also tryApart[..your fraction..]
which should get the square root out of the denominator, but even that doesn't seem to be enough. You just have a very complicated rational function with no easy integral. $\endgroup$