Unit-I LAPLACE TRANSFORM
1.5.6: Laplace Transform of Integrals:
Statement: If 𝐿{𝑓(𝑡)}=𝑓(𝑠), then 𝐿{∫𝑓(𝑡)𝑑𝑡
𝑡
0}= 𝑓
(𝑠)
𝑠 .
Proof:
𝐿{∫ 𝑓(𝑡)𝑑𝑡
𝑡
0}=∫ (𝑒−𝑠𝑡 {∫ 𝑓(𝑡)
𝑡
0𝑑𝑡})
∞
0 𝑑𝑡
By using integration by parts
𝑢 ={∫𝑓(𝑡)
𝑡
0𝑑𝑡} , 𝑣=𝑒−𝑠𝑡
={∫𝑓(𝑡)
𝑡
0𝑑𝑡∙𝑒−𝑠𝑡
−𝑠 }∞
0−∫𝑓(𝑡)
∞
0𝑒−𝑠𝑡
−𝑠 𝑑𝑡
[∵𝑑
𝑑𝑡∫ 𝑓(𝑡)
𝑡
0𝑑𝑡=𝑓(𝑡)]
= {∫ 𝑓(𝑡)
𝑡
0𝑑𝑡⋅𝑒−𝑠𝑡
−𝑠}∞
0+𝑓(𝑠)
𝑠 … …… … .(1.4)
={0−0}+𝑓(𝑠)
𝑠
=𝑓(𝑠)
𝑠 .
Example 1.7: Find the Laplace transforms of
(i) ∫𝒆−𝒕
𝒕
𝟎𝐜𝐨𝐬𝒕𝒅𝒕 (ii) ∫𝐬𝐢𝐧𝒕
𝒕
𝒕
𝟎𝒅𝒕 (iii) ∫𝒆𝒕
𝒕
𝟎𝐬𝐢𝐧𝒕
𝒕𝒅𝒕
(iv) ∫𝒆−𝒂𝒕−𝒆−𝒃𝒕
𝒕
𝒕
𝟎𝒅𝒕 (v) ∫𝐜𝐨𝐬𝒂𝒕−𝐜𝐨𝐬𝒃𝒕
𝒕
𝒕
𝟎𝒅𝒕 (vi)∫∫∫𝐜𝐨𝐬𝒂𝒕
𝒕
𝟎
𝒕
𝟎
𝒕
𝟎𝒅𝒕 𝒅𝒕 𝒅𝒕
(vii) ∫𝒕
𝒕
𝟎𝒆−𝒕𝐬𝐢𝐧𝟒𝒕𝒅𝒕 (viii) 𝒆−𝟒𝒕∫𝐬𝐢𝐧𝟑𝒕
𝒕𝒅𝒕
𝒕
𝟎 .
Solution: (i) Let 𝑓(𝑡)=cos𝑡, then
𝐿{𝑓(𝑡)}=𝐿{cos𝑡}=𝑠
𝑠2+1 =𝑓(𝑠)
∴𝐿{𝑒−𝑎𝑡𝑓(𝑡)}=𝐿{𝑒−𝑡 cos𝑡}
=𝑠+1
(𝑠+1)2+1[By First Shifting Property]
