← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Properties of Laplace transform (linearity, shift, derivative, convolution) · ODEs · asked 3× in 13 yrs · Read the full method →

Question

Given that $L\{f(t);p\}=F(p)$. Show that $\displaystyle\int_0^\infty\frac{f(t)}{t}\,dt=\int_0^\infty F(p)\,dp$. Hence evaluate the integral $\displaystyle\int_0^\infty\frac{e^{-t}-e^{-3t}}{t}\,dt$.

Technique

Fubini–Tonelli to swap the order of integration in the double integral $\int\int e^{-pt}f(t)\,dp\,dt$. The inner integral over $p$ collapses to $1/t$.

Solution

Step 1 — Prove the identity

By definition $F(p)=\int_0^\infty e^{-pt}f(t)\,dt$.

Interchange order (Fubini, valid when $f(t)/t$ is integrable on $[0,\infty)$):

(The inner integral $\int_0^\infty e^{-pt}\,dp=[-e^{-pt}/t]_0^\infty=1/t$ for $t>0$.)

Hence

Step 2 — Apply to $f(t)=e^{-t}-e^{-3t}$

$F(p)=L\{e^{-t}\}-L\{e^{-3t}\}=\dfrac{1}{p+1}-\dfrac{1}{p+3}$.

By the identity:

Antiderivative: $\ln(p+1)-\ln(p+3)=\ln\dfrac{p+1}{p+3}$.

Evaluate at limits:

• At $p=\infty$: $\ln\dfrac{p+1}{p+3}\to\ln 1=0$.
• At $p=0$: $\ln\dfrac{1}{3}=-\ln 3$.

Answer