# Limit with integral [closed]

I have this Limit: it's indeterminate. How can it be solved?


\!$$\*UnderscriptBox[\(\[Limit]$$, $$\[Epsilon] \[Rule] 1$$]\) ((1 - \[Epsilon])^2 \!$$\*SubsuperscriptBox[\(\[Integral]$$, $$0$$, $$1$$]$$\(\(-4$$\
\*SuperscriptBox[$$(1 - y)$$, $$3$$]\ $$(1 - \*FractionBox[\(25\ \((1 + \[Epsilon])$$\), $$4\ \*SqrtBox[\(6$$]\ H\ y\ $$(1 - \[Epsilon])$$\
\*SqrtBox[
FractionBox[$$1 + \[Epsilon]$$, $$1 + 11\ \[Epsilon]$$]]\)])\)\) \[DifferentialD]y\)\))/(1 + \
\[Epsilon])^2



thanks

• Is this a question about Mathematica, or about the math? Commented Mar 20 at 0:56
• math basically,
– lia
Commented Mar 20 at 7:50
• @lia Please clarify your question. The first block is product of Limit[((1-x)/(1+x))^2,x->1] (evaluates to 0) and an integral expression which only evaluates if Re[x]<=1. Commented Mar 20 at 9:28

\$Version

(* "14.0.0 for Mac OS X ARM (64-bit) (December 13, 2023)" *)

Clear["Global*"]

int = Integrate[(((1 + x*Exp[-b*y])/(1 - x*Exp[-b*y]))^2 - 1)*(1 - y)^3, {y,
0, 1}] // FullSimplify

(* -(1/(b^4))
4 (b^2 (b/(-1 + x) - 3 Log[-1 + x] + 3 Log[x]) + 6 b PolyLog[2, 1/x] +
6 PolyLog[3, 1/x] - 6 PolyLog[3, E^b/x]) *)

expr = (int /. b -> 8  Sqrt[6/d]*(f1/f2)*((1 - x)/(1 + x))) /.
d -> (1 + 11 x)/(1 + x) // FullSimplify;

Limit[((1 - x)/(1 + x))^2*expr, x -> 1]

(* Indeterminate *)


However, if f2 == f1

Limit[(((1 - x)/(1 + x))^2*expr /. f2 -> f1) // FullSimplify, x -> 1]

(* 1/256 (148 - 75 Log[5]) *)


EDIT: "Indeterminate" indicates that in general the limit does not exist. However, as Ulrich points out, one-sided limits do exist.

lim1 = (Limit[((1 - x)/(1 + x))^2*expr, x -> 1, Direction -> #] & /@
{"FromBelow", "FromAbove"}) // FullSimplify

(* {(1/(256 f1^4))
f2 (4 f1 (16 f1^2 + 18 f1 f2 + 3 f2^2) + 3 I f2 (4 f1 + f2)^2 π -
3 f2 (4 f1 + f2)^2 Log[-1 - (4 f1)/f2]), (1/(256 f1^4))
f2 (4 f1 (16 f1^2 + 18 f1 f2 + 3 f2^2) -
3 f2 (4 f1 + f2)^2 Log[1 + (4 f1)/f2])} *)


The two-sided limit exists when the one-sided limits are equal

cond = Equal @@ lim1 // FullSimplify

(* (f2 (4 f1 + f2) (I π - Log[-1 - (4 f1)/f2] + Log[1 + (4 f1)/f2]))/f1 == 0 *)


For example,

cond /. f2 -> f1

(* True *)

• could you explain that? it seems it does not answer my question
– lia
Commented Mar 20 at 8:15

Try

int = Integrate[(((1 + x*Exp[-b*y])/(1 - x*Exp[-b*y]))^2 -1)*(1 - y)^3, {y, 0, 1} ]
(*(4 (-((b^3 x)/(-1 + x)) + 3 b^2 Log[1 - x] + 6 b PolyLog[2, x] -6 PolyLog[3, x] + 6 PolyLog[3, E^-b x]))/b^4*)

limit=Limit[Simplify[((1 - x)/(1 + x))^2 int /.b -> 8 Sqrt[6/d]*(f1/f2)*((1 - x)/(1 + x)) /.d -> (1 + 11 x)/(1 + x)], x -> 1, Direction -> "FromBelow"]

(*(f2 (4 f1 (16 f1^2 + 18 f1 f2 + 3 f2^2) -3 f2 (4 f1 + f2)^2 Log[1 + (4 f1)/f2]))/(256 f1^4)*)


plot result

Plot3D[limit, {f1, -1, 1}, {f2, -1, 1}, MaxRecursion -> 3,PlotPoints -> 100 , AxesLabel -> {f1, f2, "limit"}, MeshFunctions -> (#3 &)]
`