Examples for Section 7.6
Improper Integrals

p. 603

1.

(∫_0^b)\[DifferentialD](x/((x^2) + 1))

Limit[ArcTan[b],  b -> ∞]

(∫_0^∞)\[DifferentialD](x/((x^2) + 1))

5.

\(TraditionalForm
    \`∫\_\(-1\)\%1\( 1\/x\^\(2/3\)\) dx\  = \ 
      \(∫\_\(-1\)\%0\( x\^\(\(-2\)/3\)\) dx\  + \ 
          ∫\_0\%1\( x\^\(\(-2\)/3\)\) dx\  = \ 
        \(3\  + \ 3\  = \ 6\)\)\)

(∫_(-1)^b) (x^(-2/3)) \[DifferentialD]x

Limit[-3 (((-1))^(1/3)) + 3 (b^(1/3)),  b -> 0]

This is actually 3.

(∫_a^1) (x^(-2/3)) \[DifferentialD]x

Limit[3 - 3 (a^(1/3)),  a -> 0]

9.

(∫_a^(-2))(2/((x^2) - 1)) \[DifferentialD]x

Limit[Log[3] - Log[-1 + a] + Log[1 + a],  
      a -> -∞]

(∫_(-∞)^(-2))(2/((x^2) - 1)) 
        \[DifferentialD]x

N[%]

N[Log[3]]

21.

(∫_a^0) θ (E^θ) 
        \[DifferentialD]θ

Limit[-1 - (-1 + a) (E^a),  a -> -∞]

(∫_(-∞)^0) θ (E^θ) 
        \[DifferentialD]θ

22. (Modified)

\(TraditionalForm
      \`∫\_0\%∞\( e\^\(-θ\)\) \(cos(θ)\) 
          dθ\)

(∫_0^b) (E^(-θ)) Cos[θ] 
        \[DifferentialD]θ

Limit[(1/2) - (1/2) (E^(-b)) (Cos[b] - Sin[b]),  
      b -> ∞]

We can see that cos(b) - sin(b) is never greater than 2 or less that -2, but (e^(-b)) -> 0 as b -> ∞. Therefore the limit is (1/2). Oddly enough, Mathematica can calculate the improper integral.

(∫_0^∞) (E^(-θ)) Cos[θ] 
        \[DifferentialD]θ

25.

(∫_a^1) x Log[x] \[DifferentialD]x

Limit[-(1/4) - (1/4) (a^2) (-1 + 2 Log[a]),  
      a -> 0]

(∫_0^1) x Log[x] \[DifferentialD]x

49.

\(TraditionalForm
      \`∫\_2\%∞\( 1\/\@\(v - 1\)\) dv\)
Use the Limit Comparison test: (1/(Sqrt[v - 1])) grows at the same rate as (1/(Sqrt[v])).

Limit[(1/(Sqrt[v - 1]))/(1/(Sqrt[v])),  v -> ∞]

The integral \(TraditionalForm
      \`∫\_2\%∞\( 1\/\@v\) dv\) diverges:

(∫_2^b) (1/(Sqrt[v])) \[DifferentialD]v

Limit[-2 (Sqrt[2]) + 2 (Sqrt[b]),  b -> ∞]

Therefore, the integral \(TraditionalForm
      \`∫\_2\%∞\( 1\/\@\(v - 1\)\) dv\) diverges.

53.

\(TraditionalForm
      \`∫\_1\%∞\(\@\( x + 1\)\/x\^2\) 
          dx\)
Use the Limit Comparison Test: ((Sqrt[x + 1])/(x^2))grows at the same rate as ((Sqrt[x])/(x^2)) =  (x^(-3/2)).

Limit[((Sqrt[x + 1])/(x^2))/(x^(-3/2)), x -> ∞]

The integral \(TraditionalForm
      \`∫\_1\%∞\( x\^\(\(-3\)/2\)\) 
          dx\) converges:

(∫_1^b) (x^(-3/2)) \[DifferentialD]x

Limit[2 - (2/(Sqrt[b])),  b -> ∞]

Therefore, the integral \(TraditionalForm
      \`∫\_1\%∞\(\@\( x + 1\)\/x\^2\) 
          dx\) converges.

62.

\(TraditionalForm
      \`∫\_\(-∞\)\%∞dx
              \/\@\(x\^4 + 1\)\  = \ 
        ∫\_\(-∞\)\%0dx\/\@\(x\^4 + 1\) + 
          ∫\_0\%∞dx\/\@\(x\^4 + 1\)\)
Use the Limit Comparison Test: (1/(Sqrt[(x^4) + 1]))grows at the same rate as (1/(x^2)):

Limit[(1/(Sqrt[(x^4) + 1]))/(1/(x^2)),  x -> ∞]

The integral \(TraditionalForm
      \`∫\_1\%∞\( 1\/x\^2\) dx\) converges,

(∫_1^a) (x^(-2)) \[DifferentialD]x

Limit[1 - (1/a),  a -> ∞]

Therefore, the integral \(TraditionalForm
      \`∫\_0\%∞dx\/\@\(x\^4 + 1\)\)converges. A similar argument proves that the integral \(TraditionalForm
      \`∫\_\(-∞\)\%0
          dx\/\@\(x\^4 + 1\)\) converges. Therefore the original integral \(TraditionalForm
      \`∫\_\(-∞\)\%∞dx
            \/\@\(x\^4 + 1\)\) converges.

91.

a)

Si[x_]  :=  (∫_0^x) Sin[t]/t \[DifferentialD]t

Plot[ Sin[t]/t, {t, 0, 6  π},  PlotRange -> {-0.5, 1}]

Since sin(t)/tchanges sign repeatedly, the integral repeatedly adds and subtracts the area between the graph and the x-axis as x increases. This suggests that the function Si(x) will also oscillate.

Plot[Si[x],  {x, 0, 25},  PlotRange -> {0, 2}]

b)

Limit[Si[x],  x -> ∞]

N[Si[50]]

N[Si[100]]

N[Si[1000]]

N[π/2]

It looks like the limit is π/2.