I am new to Mathematica, and I am trying to simplify this expression when Y=1
and Y1 -> ∞
X=2 * Pi * Y^2 * Y1 * Integrate[1/((Y^2 + s)^2 * Sqrt[(Y1^2 + s)]),{s,0,∞}]
Fixing syntax errors and adding assumptions on your variables:
Assuming[Y > 0 && Y1 > 0,
X = 2 π Y^2 Y1 Integrate[1/((Y^2+s)^2*Sqrt[(Y1^2+s)]), {s,0,∞}]
];
X //TeXForm
$\frac{2 \pi Y \operatorname{Y1} \left(Y \cos ^{-1}\left(\frac{\operatorname{Y1}}{Y}\right)-\operatorname{Y1} \sqrt{1-\frac{\operatorname{Y1}^2}{Y^2}}\right)}{\left(Y^2-\operatorname{Y1}^2\right)^{3/2}}$
Then, replace Y
with 1
and take the limit as Y1
goes to $\infty$:
Limit[X /. Y->1, Y1->∞]
2 π
Update
If you're interested in the Y
and Y1
dependence for large Y1
, you could do:
Series[X,{Y1,Infinity,2},Assumptions->Y>0] //TeXForm
$2 \pi +\frac{2 \pi Y^2}{\operatorname{Y1}^2}+O\left(\left(\frac{1}{\operatorname{Y1}}\right)^3\right)$
A new EXPERIMENTAL function in version 11.3 is AsymptoticIntegrate
$Version
"11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)"
X = 2*Pi*Y^2*Y1*
AsymptoticIntegrate[
1/((Y^2 + s)^2*Sqrt[(Y1^2 + s)]), {s, 0, ∞}, {Y1, Infinity, 1}]
(* 2 π *)
X = 2*Pi*Y^2*Y1*
AsymptoticIntegrate[
1/((Y^2 + s)^2*Sqrt[(Y1^2 + s)]), {s, 0, ∞}, {Y1, Infinity,
3}] // FullSimplify
(* 2 π (1 + Y^2/Y1^2) *)
Quite simple. As $Y=1$, then $X$ simplifies to:
$Version
11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)
X=2*Pi*Y1*Integrate[1/((1 + s)^2*Sqrt[Y1^2 + s]), {s, 0, Infinity}]
$$\text{ConditionalExpression}\left[\frac{\pi \text{Y1} \sqrt{\text{Y1}^2} \left(\pi \sqrt{\frac{1}{\text{Y1}^2}}-2 \sqrt{1-\text{Y1}^2}-\frac{2 \sin ^{-1}(\text{Y1})}{\text{Y1}}\right)}{\left(1-\text{Y1}^2\right)^{3/2}},\Re\left(\text{Y1}^2\right)>0\right]$$
Now, taking limit when $Y1\rightarrow\infty$:
Limit[X, Y1 -> \[Infinity]]
$2\pi$