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UPSC 2026 Maths Optional Paper 1 Q7c-ii — Step-by-Step Solution

10 marks · Section B

Properties of Laplace transform (linearity, shift, derivative, convolution) · ODEs · asked 4× in 14 yrs · Read the full method →

Question

If $L\{f(t)\} = \bar{f}(s)$, then show that $L\left\{\dfrac{f(t)}{t}\right\} = \displaystyle\int_{s}^{\infty} \bar{f}(u)\,du$. Hence evaluate $L\left\{\dfrac{\cos at - \cos bt}{t}\right\}$.

Technique

Write $\int_s^\infty \bar f(u)\,du$ as a double integral, interchange the order of integration (justified by Fubini–Tonelli under the exponential-order hypothesis), and do the inner $u$-integral first — it produces the factor $1/t$. Then apply the result to $f(t) = \cos at - \cos bt$.

Solution

The transform of $f(t)/t$

Hypotheses. Let $f$ be piecewise continuous on $[0,\infty)$ and of exponential order $\gamma$, i.e. $|f(t)| \le Ke^{\gamma t}$, so that $\bar f(u) = \displaystyle\int_0^\infty e^{-ut}f(t)\,dt$ converges for $u > \gamma$. Assume further that $\displaystyle\lim_{t\to 0^+}\frac{f(t)}{t}$ exists (finite), so $f(t)/t$ is itself piecewise continuous on $[0,\infty)$ and of exponential order, and $L\{f(t)/t\}$ exists for $s > \gamma$. Fix $s > \gamma$.

The double integral. By definition,

Justifying the interchange. On the quarter-plane $\{t \ge 0,\ u \ge s\}$,

the last integral being finite precisely because $|f(t)|/t$ is bounded near $t=0$ and of exponential order $\gamma < s$. The double integral is therefore absolutely convergent, and Fubini’s theorem permits the interchange.

Interchanging.

The inner integral, for $t > 0$, is

Hence

Application: $f(t) = \cos at - \cos bt$

This $f$ qualifies: it is continuous, bounded (exponential order $\gamma = 0$), and

so $f(t)/t \to 0$ as $t\to 0^+$ — the removable singularity condition holds. (This is exactly why the $-\cos bt$ is there; $L\{\cos at/t\}$ alone does not exist.)

Using $L\{\cos kt\} = \dfrac{s}{s^2+k^2}$,

Therefore, for $s > 0$,

As $u \to \infty$, $\dfrac{u^2+a^2}{u^2+b^2} \to 1$, so the upper limit contributes $\log 1 = 0$. (Note the two logarithms must be combined before taking the limit — separately each diverges; this is the step that carries the marks.) Hence

Answer