Metamath Proof Explorer


Theorem itgcl

Description: The integral of an integrable function is a complex number. This is Metamath 100 proof #86. (Contributed by Mario Carneiro, 29-Jun-2014)

Ref Expression
Hypotheses itgmpt.1 ( ( 𝜑𝑥𝐴 ) → 𝐵𝑉 )
itgcl.2 ( 𝜑 → ( 𝑥𝐴𝐵 ) ∈ 𝐿1 )
Assertion itgcl ( 𝜑 → ∫ 𝐴 𝐵 d 𝑥 ∈ ℂ )

Proof

Step Hyp Ref Expression
1 itgmpt.1 ( ( 𝜑𝑥𝐴 ) → 𝐵𝑉 )
2 itgcl.2 ( 𝜑 → ( 𝑥𝐴𝐵 ) ∈ 𝐿1 )
3 eqid ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) = ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) )
4 3 dfitg 𝐴 𝐵 d 𝑥 = Σ 𝑘 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ if ( ( 𝑥𝐴 ∧ 0 ≤ ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) ) , ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) , 0 ) ) ) )
5 fzfid ( 𝜑 → ( 0 ... 3 ) ∈ Fin )
6 ax-icn i ∈ ℂ
7 elfznn0 ( 𝑘 ∈ ( 0 ... 3 ) → 𝑘 ∈ ℕ0 )
8 7 adantl ( ( 𝜑𝑘 ∈ ( 0 ... 3 ) ) → 𝑘 ∈ ℕ0 )
9 expcl ( ( i ∈ ℂ ∧ 𝑘 ∈ ℕ0 ) → ( i ↑ 𝑘 ) ∈ ℂ )
10 6 8 9 sylancr ( ( 𝜑𝑘 ∈ ( 0 ... 3 ) ) → ( i ↑ 𝑘 ) ∈ ℂ )
11 elfzelz ( 𝑘 ∈ ( 0 ... 3 ) → 𝑘 ∈ ℤ )
12 eqidd ( 𝜑 → ( 𝑥 ∈ ℝ ↦ if ( ( 𝑥𝐴 ∧ 0 ≤ ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) ) , ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) , 0 ) ) = ( 𝑥 ∈ ℝ ↦ if ( ( 𝑥𝐴 ∧ 0 ≤ ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) ) , ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) , 0 ) ) )
13 eqidd ( ( 𝜑𝑥𝐴 ) → ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) = ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) )
14 12 13 2 1 iblitg ( ( 𝜑𝑘 ∈ ℤ ) → ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ if ( ( 𝑥𝐴 ∧ 0 ≤ ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) ) , ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) , 0 ) ) ) ∈ ℝ )
15 11 14 sylan2 ( ( 𝜑𝑘 ∈ ( 0 ... 3 ) ) → ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ if ( ( 𝑥𝐴 ∧ 0 ≤ ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) ) , ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) , 0 ) ) ) ∈ ℝ )
16 15 recnd ( ( 𝜑𝑘 ∈ ( 0 ... 3 ) ) → ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ if ( ( 𝑥𝐴 ∧ 0 ≤ ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) ) , ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) , 0 ) ) ) ∈ ℂ )
17 10 16 mulcld ( ( 𝜑𝑘 ∈ ( 0 ... 3 ) ) → ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ if ( ( 𝑥𝐴 ∧ 0 ≤ ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) ) , ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) , 0 ) ) ) ) ∈ ℂ )
18 5 17 fsumcl ( 𝜑 → Σ 𝑘 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ if ( ( 𝑥𝐴 ∧ 0 ≤ ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) ) , ( ℜ ‘ ( 𝐵 / ( i ↑ 𝑘 ) ) ) , 0 ) ) ) ) ∈ ℂ )
19 4 18 eqeltrid ( 𝜑 → ∫ 𝐴 𝐵 d 𝑥 ∈ ℂ )