Laplace transform

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2018, 2019)
• Priority tier: T3
• Marks (count): 10 (1), 5 (1)
• Average solve time: ~6 min
• Difficulty mix: easy 2
• Section: B | Dominant type: proof

Why This Chapter Matters

The Laplace-transform definition questions in UPSC are uniformly easy: both exam appearances reduce to one master formula $\mathcal{L}\{t^p\}=\Gamma(p+1)/s^{p+1}$ with a handful of Gamma values plugged in. Knowing $\Gamma(1/2)=\sqrt\pi$, $\Gamma(3/2)=\frac{1}{2}\sqrt\pi$, and the recursion $\Gamma(z+1)=z\Gamma(z)$ covers every question that has appeared and every plausible variant. This topic is also the entry point for P1-OD-15 (properties) and P1-OD-16 (inverse), so the Gamma-function fluency built here pays dividends across the entire Laplace block.

Minimum Theory

Definition. The Laplace transform of $f(t)$ is

$\mathcal{L}\{f\}(s) = \int_0^\infty e^{-st}f(t)\,dt,$

defined for those $s>0$ for which the integral converges. The transform is linear: $\mathcal{L}\{af+bg\}=a\mathcal{L}\{f\}+b\mathcal{L}\{g\}$.

Power function transform. For $p>-1$ and $s>0$, substituting $u=st$ in the defining integral:

$\mathcal{L}\{t^p\} = \int_0^\infty e^{-st}t^p\,dt = \frac{1}{s^{p+1}}\int_0^\infty e^{-u}u^p\,du = \frac{\Gamma(p+1)}{s^{p+1}}.$

For non-negative integers $p=n$: $\Gamma(n+1)=n!$, giving the familiar $\mathcal{L}\{t^n\}=n!/s^{n+1}$.

Key Gamma values. $\Gamma(1/2)=\sqrt\pi$. Recursion: $\Gamma(z+1)=z\Gamma(z)$, so $\Gamma(3/2)=\frac{1}{2}\Gamma(1/2)=\frac{\sqrt\pi}{2}$; $\Gamma(5/2)=\frac{3}{4}\sqrt\pi$; $\Gamma(n+3/2)=\frac{(2n+1)!}{2^{2n+1}\,n!}\sqrt\pi$. The integral exists even when $f(t)\to\infty$ as $t\to0^+$, provided $p>-1$ (the singularity is integrable).

Question Archetypes

transform-computation (2 question(s); 2018, 2019)

Recognition Cues

• “Find the Laplace transform of $t^{-1/2}$” or $t^{1/2}$, or $1/\sqrt t$.
• “Prove that $\mathcal{L}\{t^{n+1/2}\}=\Gamma(n+3/2)/s^{n+3/2}$.”
• Any fractional power of $t$ — the naive formula $n!/s^{n+1}$ does not apply; use the Gamma function.

Solution Template

1. State the master formula $\mathcal{L}\{t^p\}=\Gamma(p+1)/s^{p+1}$ for $p>-1$.
2. Derive it (if proof is asked): substitute $u=st$ to convert $\int_0^\infty e^{-st}t^p\,dt$ into $\Gamma(p+1)/s^{p+1}$.
3. Plug in the specific value of $p$.
4. Evaluate the Gamma value using $\Gamma(1/2)=\sqrt\pi$ and the recursion.
5. Box the answer with the constraint $s>0$.

Worked Example

2018 Paper 1, 2018-P1-Q5d-i (5 marks)

Find the Laplace transform of $f(t)=\dfrac{1}{\sqrt t}$.

Step 1 — Identify. $f(t)=t^{-1/2}$, so $p=-1/2$. Check $p>-1$: $-1/2>-1$ ✓, so the transform exists.

Step 2 — Apply the formula.

$\mathcal{L}\{t^{-1/2}\} = \frac{\Gamma(-1/2+1)}{s^{-1/2+1}} = \frac{\Gamma(1/2)}{s^{1/2}}.$

Step 3 — Use $\Gamma(1/2)=\sqrt\pi$.

$\boxed{\;\mathcal{L}\!\left\{\frac{1}{\sqrt t}\right\} = \sqrt{\frac{\pi}{s}},\quad s>0.\;}$

Direct derivation (if asked to derive from scratch): Substitute $t=u^2/s$ ($dt=2u\,du/s$):

$\int_0^\infty e^{-st}t^{-1/2}\,dt = \int_0^\infty e^{-u^2}\frac{\sqrt s}{u}\cdot\frac{2u\,du}{s} = \frac{2}{\sqrt s}\int_0^\infty e^{-u^2}\,du = \frac{2}{\sqrt s}\cdot\frac{\sqrt\pi}{2} = \sqrt{\frac\pi s}.$

2019 Paper 1, 2019-P1-Q6c-ii (10 marks)

Find the Laplace transforms of $t^{-1/2}$ and $t^{1/2}$. Prove that $\mathcal{L}\{t^{n+1/2}\}=\Gamma(n+3/2)/s^{n+3/2}$ for $n\in\mathbb{N}$.

Step 1 — Master formula derivation. For $p>-1$, substitute $u=st$:

$\mathcal{L}\{t^p\} = \int_0^\infty e^{-st}t^p\,dt = \frac{1}{s^{p+1}}\int_0^\infty e^{-u}u^p\,du = \frac{\Gamma(p+1)}{s^{p+1}}.$

Step 2 — Transform of $t^{-1/2}$. $p=-1/2$:

$\mathcal{L}\{t^{-1/2}\} = \frac{\Gamma(1/2)}{s^{1/2}} = \boxed{\sqrt{\frac{\pi}{s}}}.$

Step 3 — Transform of $t^{1/2}$. $p=1/2$; use $\Gamma(3/2)=\frac{1}{2}\Gamma(1/2)=\frac{\sqrt\pi}{2}$:

$\mathcal{L}\{t^{1/2}\} = \frac{\Gamma(3/2)}{s^{3/2}} = \boxed{\frac{\sqrt\pi}{2s^{3/2}}}.$

Step 4 — General formula for $t^{n+1/2}$. Take $p=n+1/2$ (so $p+1=n+3/2$):

$\mathcal{L}\{t^{n+1/2}\} = \frac{\Gamma(n+1/2+1)}{s^{n+1/2+1}} = \frac{\Gamma(n+3/2)}{s^{n+3/2}}.\qquad\blacksquare$

$\boxed{\;\mathcal{L}\{t^{n+1/2}\} = \frac{\Gamma(n+3/2)}{s^{n+3/2}},\quad s>0,\quad n\in\mathbb{N}.\;}$

Common Traps

• The formula $\mathcal{L}\{t^n\}=n!/s^{n+1}$ applies only to non-negative integers. For $t^{-1/2}$, $t^{1/2}$, or any $t^{n+1/2}$, use the Gamma function version.
• The condition for convergence is $p>-1$, not $p\geq0$. The transform of $t^{-1/2}$ exists because $-1/2>-1$, even though the integrand has a singularity at $t=0$.
• $\Gamma(3/2)=\frac{1}{2}\sqrt\pi$, not $\sqrt\pi$. Writing $\Gamma(3/2)=\sqrt\pi$ gives the wrong answer for $\mathcal{L}\{t^{1/2}\}$.
• When asked to “prove” the formula for $t^{n+1/2}$, derive the master formula $\mathcal{L}\{t^p\}=\Gamma(p+1)/s^{p+1}$ by the $u=st$ substitution — do not just state it without justification.

Marks-Aware Writing

5-mark question (2018): State the formula, identify $p=-1/2$, use $\Gamma(1/2)=\sqrt\pi$, box the answer. Four clear lines earn all 5 marks. If you derive the result from the Gaussian integral directly (the $t=u^2/s$ substitution), that also earns full marks.

10-mark question (2019): Three sub-results: $\mathcal{L}\{t^{-1/2}\}$ (3 marks), $\mathcal{L}\{t^{1/2}\}$ (3 marks), general formula proof (4 marks). The derivation of the master formula via $u=st$ is the proof core — state it once and apply it three times. The general-case proof is essentially a one-line substitution $p=n+1/2$.

Practice Set

(All UPSC questions on this atom are fully worked above.)