Fourier and Laplace Methods Questions and Answers

Q: The function \(f(x) = \begin{cases} x, & -\pi \lt x \lt 0 \\ -x, & 0 \lt x \lt \pi \end{cases}\) is expanded as a Fourier series of the form \(a_{0} + \sum_{n=1}^{\infty} a_{n}\cos(nx) + \sum_{n=1}^{\infty} b_{n}\sin(nx)\). Which of the following is true?

The answer is \(a_{0} \neq 0, b_{n} = 0\).The function equals \(-|x|\): it is even, since \(f(-x) = f(x)\). An even function has no sine terms, so every \(b_{n} = 0\).The constant term is the mean value over the period:\[ \begin{aligned} a_{0} &= \frac{1}{2\pi}\int_{-\pi}^{\pi} (-|x|)\,dx \\ &= -\frac{\pi}{2} \neq 0 \end{aligned} \]Treating the function as odd (because each piece is linear in x) would wrongly give \(a_{0} = 0\) and nonzero \(b_{n}\).

Q: A piecewise regular function \(f(x)\) is shown in the figure. It is expanded in Fourier series given by\[ \begin{aligned} f(x) = \frac{a_{0}}{2} &+ \sum_{n=1}^{\infty} a_{n}\cos(nx) \\ &+ \sum_{n=1}^{\infty} b_{n}\sin(nx), \end{aligned} \]where \(a_{0}, a_{n}, b_{n}\)’s are the Fourier coefficients.Which of the following options is correct?

The correct option is that all \(a_n\) are zero.On \(0 < x < \pi\) the graph is \(f(x) = \frac{\pi}{2} - x\), repeated with period \(\pi\). Shifting one period back, on \(-\pi < x < 0\) it is \(f(x) = -\frac{\pi}{2} - x\). So for \(0 < x < \pi\):\[ f(-x) = -\tfrac{\pi}{2} + x = -f(x) \]The function is odd. The cosine coefficients \(a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx\,dx\) integrate an odd function over a symmetric interval, so every one of them, \(a_0\) included, is zero; only sine terms survive.Since \(a_0\) is twice the average value and the sawtooth averages to zero, \(a_0 = 3\pi\) and \(a_0 = 3\pi/2\) are wrong. The \(b_n\) cannot all vanish, or \(f\) would be identically zero.

Q: The inverse Fourier transform of \(e^{-|k|}\) is :

The inverse transform is \(\frac{1}{\pi}\frac{1}{1+x^{2}}\).With the convention \(f(x) = \frac{1}{2\pi}\int F(k) e^{ikx} dk\), split the integral at \(k = 0\):\[ \begin{aligned} f(x) &= \frac{1}{2\pi}\Big[\int_{0}^{\infty} e^{-k(1-ix)}dk + \int_{0}^{\infty} e^{-k(1+ix)}dk\Big] \\ &= \frac{1}{2\pi}\Big[\frac{1}{1-ix} + \frac{1}{1+ix}\Big] \\ &= \frac{1}{2\pi}\cdot\frac{2}{1+x^{2}} = \frac{1}{\pi}\frac{1}{1+x^{2}} \end{aligned} \]This is the Lorentzian (Cauchy) profile. A Gaussian \(e^{-x^{2}}\) is the transform of a Gaussian, not of \(e^{-|k|}\); \(e^{-|x|}\) simply copies the input; and \(1/(1+x)\) is not even an even function, while the transform of an even function must be even.

Q: Given that Fourier integral representation for the function\[ f(x) = \begin{cases} 0 & \text{if } |x| \gt 1 \\ 1 & \text{if } |x| \lt 1 \end{cases} \]is \( f(x)=\frac{2}{\pi}\int_0^\infty\frac{\cos\omega x\sin\omega}{\omega}d\omega \)Which of the following options is correct ?

The correct statement is the piecewise result: the integral equals \( \pi/2 \) for \( 0 \le x \lt 1 \), \( \pi/4 \) at \( x = 1 \) and 0 for \( x \gt 1 \).Rearranging the given representation, the integral is \( \frac{\pi}{2} f(x) \) wherever \( f \) is continuous:\[ \begin{aligned} &\int_0^\infty \frac{\cos\omega x \sin\omega}{\omega}\, d\omega \\ &= \frac{\pi}{2} f(x) \end{aligned} \]So it is \( \pi/2 \) inside \( |x| \lt 1 \) and 0 outside. At the jump \( x = 1 \) a Fourier integral converges to the mean of the left and right limits, \( \frac{\pi}{2} \cdot \frac{1 + 0}{2} = \frac{\pi}{4} \).Putting \( x = 0 \) gives the Dirichlet integral \( \int_{0}^{\infty} \frac{\sin \omega}{\omega} d\omega = \frac{\pi}{2} \), so the options giving \( \pi \) and \( \pi/4 \) are wrong, and the integral is 0 only for \( |x| \gt 1 \), not for every \( x \).

Q: Fourier series for the function\[ f(\theta) = \begin{cases} +1 & 0 \lt \theta \lt \pi \\ -1 & \pi \lt \theta \lt 2\pi \end{cases} \]there are :

The series contains odd sine terms only.The square wave is an odd function about \(\theta = 0\) (shifting the second half to \(-\pi \lt \theta \lt 0\) gives \(f(-\theta) = -f(\theta)\)), so every cosine coefficient and the constant term vanish. For the sine coefficients:\[ \begin{aligned} b_{n} &= \frac{2}{\pi}\int_{0}^{\pi}\sin n\theta\,d\theta \\ &= \frac{2}{n\pi}(1 - \cos n\pi) \end{aligned} \]This is \(4/n\pi\) for odd n and 0 for even n, giving \(f = \frac{4}{\pi}(\sin\theta + \frac{1}{3}\sin 3\theta + \dots)\). 'Sine terms only' is incomplete because the even harmonics are absent, and the cosine options fail the odd symmetry.

Q: Consider the function\(f(x) = |x| \quad -1 \le x \le 1\)Then the Fourier series for \(f\) is :

The Fourier series is \(\frac{1}{2} - \frac{4}{\pi^{2}}(\cos \pi x + \frac{1}{9}\cos 3\pi x + \cdots)\), the cosine series with odd harmonics only.\(|x|\) is even, so it has a cosine series only (which rules out the options with sine terms). On [-1, 1], with period 2:\[ \begin{aligned} a_{0} &= \int_{-1}^{1} |x|\,dx = 1 \\ a_{n} &= 2\int_{0}^{1} x\cos n\pi x\,dx \\ &= \frac{2((-1)^{n} - 1)}{n^{2}\pi^{2}} \end{aligned} \]So \(a_{n} = -4/(n^{2}\pi^{2})\) for odd \(n\) and 0 for even \(n\), and the constant term is \(a_{0}/2 = 1/2\). The minus sign is right: at \(x = 0\) the series must give 0, and \(1/2 - (4/\pi^{2})(\pi^{2}/8) = 0\).

Q: Laplace transform of \(J_{0}(a\, t)\) where \(J_{0}\) denotes the Bessel function of zero order is :

The Laplace transform is \(\dfrac{1}{\sqrt{s^{2} + a^{2}}}\).Using the series \(J_{0}(at) = \sum (-1)^{k}(at/2)^{2k}/(k!)^{2}\) and \(L\{t^{2k}\} = (2k)!/s^{2k+1}\), the transform sums to\[ \begin{aligned} &\frac{1}{s}\left(1 + \frac{a^{2}}{s^{2}}\right)^{-1/2} \\ &= \frac{1}{\sqrt{s^{2} + a^{2}}} \end{aligned} \]by the binomial series. \(1/(s^{2} + a^{2})\) is (up to a factor \(a\)) the transform of \(\sin at\), and the forms with \(s^{2} - a^{2}\) belong to the modified Bessel function \(I_{0}(at)\) and to \(\sinh at\).

Q: Fourier cosine transform of \(f(t) = e^{-t^{2}}\) is :

The Fourier cosine transform is \(\dfrac{1}{\sqrt{2}}e^{-s^{2}/4}\).With the convention \(F_{c}(s) = \sqrt{2/\pi}\int_{0}^{\infty} f(t)\cos st\,dt\), and the standard integral\[ \begin{aligned} \int_{0}^{\infty} e^{-t^{2}}\cos st\,dt &= \frac{\sqrt{\pi}}{2}e^{-s^{2}/4} \\ F_{c}(s) &= \sqrt{\frac{2}{\pi}}\cdot\frac{\sqrt{\pi}}{2}e^{-s^{2}/4} = \frac{1}{\sqrt{2}}e^{-s^{2}/4} \end{aligned} \]The coefficient \(1/\sqrt{2}\) in every option shows this convention. The exponent must be \(-s^{2}/4\); the other exponents come from mis-scaling the Gaussian (\(-s^{2}/2\) is the transform of \(e^{-t^{2}/2}\)).

Q: If \(L\{f(t)\} = F(s)\) and \(L\{g(t)\} = G(s)\), then \(L^{-1}\{(F(s)G(s)\}\) is :

\(L^{-1}\{F(s)G(s)\}\) is \(\int_{0}^{t} f(x)\,g(t - x)\,dx\).This is the convolution theorem: the Laplace transform of the convolution \((f * g)(t) = \int_{0}^{t} f(x)g(t - x)\,dx\) is the product \(F(s)G(s)\). It is proved by changing the order of integration in \(\int_{0}^{\infty} e^{-st}\int_{0}^{t} f(x)g(t - x)\,dx\,dt\) and substituting \(\tau = t - x\).The inverse transform of a product is not the integral of the product \(f(x)g(x)\), and the arguments \(t + x\) or \(x - t\) in the other options do not give a convolution.

Q: The complex Fourier transform of \(e^{-|x|}\) is :

The complex Fourier transform is \(\frac{2}{1 + n^{2}}\).Using \(F(n) = \int_{-\infty}^{\infty} f(x)e^{inx}dx\) and splitting the range at zero, where \(|x|\) changes sign:\[ \begin{aligned} F(n) &= \int_{-\infty}^{0} e^{(1 + in)x}dx \\ &\quad + \int_{0}^{\infty} e^{-(1 - in)x}dx \\ &= \frac{1}{1 + in} + \frac{1}{1 - in} \\ &= \frac{2}{1 + n^{2}} \end{aligned} \]The paper uses the convention without the factor \(1/\sqrt{2\pi}\). With it, the result would be \(\sqrt{2/\pi}/(1 + n^{2})\), which is not offered. The other numerators come from dropping one of the two halves (1) or mishandling the constants (4 or \(\pi\)).

The function \(f(x) = \begin{cases} x, & -\pi \lt x \lt 0 \\ -x, & 0 \lt x \lt \pi \end{cases}\) is expanded as a Fourier series of the form \(a_{0} + \sum_{n=1}^{\infty} a_{n}\cos(nx) + \sum_{n=1}^{\infty} b_{n}\sin(nx)\). Which of the following is true?

A.\(a_{0} \neq 0, b_{n} = 0\)
B.\(a_{0} \neq 0, b_{n} \neq 0\)
C.\(a_{0} = 0, b_{n} = 0\)
D.\(a_{0} = 0, b_{n} \neq 0\)

Exam Year(s)

IIT JAM - 2018,

Answer: A

The answer is \(a_{0} \neq 0, b_{n} = 0\).

The function equals \(-|x|\): it is even, since \(f(-x) = f(x)\). An even function has no sine terms, so every \(b_{n} = 0\).

The constant term is the mean value over the period:

\[ \begin{aligned} a_{0} &= \frac{1}{2\pi}\int_{-\pi}^{\pi} (-|x|)\,dx \\ &= -\frac{\pi}{2} \neq 0 \end{aligned} \]

Treating the function as odd (because each piece is linear in x) would wrongly give \(a_{0} = 0\) and nonzero \(b_{n}\).

Discuss in Forum

A piecewise regular function \(f(x)\) is shown in the figure. It is expanded in Fourier series given by

\[ \begin{aligned} f(x) = \frac{a_{0}}{2} &+ \sum_{n=1}^{\infty} a_{n}\cos(nx) \\ &+ \sum_{n=1}^{\infty} b_{n}\sin(nx), \end{aligned} \]

where \(a_{0}, a_{n}, b_{n}\)’s are the Fourier coefficients.

Sawtooth y against x: from x = 0 to pi the line falls from pi/2 through zero at pi/2 to -pi/2, jumps back to pi/2 at x = pi, and the same line repeats from pi to 2 pi, crossing zero at 3 pi/2

Which of the following options is correct?

A.\(a_{0} = 3\pi\)
B.All \(a_{n}\) are zero
C.All \(b_{n}\) are zero
D.\(a_{0} = \frac{3\pi}{2}\)

Exam Year(s)

IIT JAM - 2026,

Answer: B

The correct option is that all \(a_n\) are zero.

On \(0 < x < \pi\) the graph is \(f(x) = \frac{\pi}{2} - x\), repeated with period \(\pi\). Shifting one period back, on \(-\pi < x < 0\) it is \(f(x) = -\frac{\pi}{2} - x\). So for \(0 < x < \pi\):

\[ f(-x) = -\tfrac{\pi}{2} + x = -f(x) \]

The function is odd. The cosine coefficients \(a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx\,dx\) integrate an odd function over a symmetric interval, so every one of them, \(a_0\) included, is zero; only sine terms survive.

Since \(a_0\) is twice the average value and the sawtooth averages to zero, \(a_0 = 3\pi\) and \(a_0 = 3\pi/2\) are wrong. The \(b_n\) cannot all vanish, or \(f\) would be identically zero.

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The inverse Fourier transform of \(e^{-|k|}\) is :

A.\(\frac{1}{\pi} \frac{1}{1+x^{2}}\)
B.\(e^{-x^{2}}\)
C.\(e^{-|x|}\)
D.\(\frac{1}{1+x}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The inverse transform is \(\frac{1}{\pi}\frac{1}{1+x^{2}}\).

With the convention \(f(x) = \frac{1}{2\pi}\int F(k) e^{ikx} dk\), split the integral at \(k = 0\):

\[ \begin{aligned} f(x) &= \frac{1}{2\pi}\Big[\int_{0}^{\infty} e^{-k(1-ix)}dk + \int_{0}^{\infty} e^{-k(1+ix)}dk\Big] \\ &= \frac{1}{2\pi}\Big[\frac{1}{1-ix} + \frac{1}{1+ix}\Big] \\ &= \frac{1}{2\pi}\cdot\frac{2}{1+x^{2}} = \frac{1}{\pi}\frac{1}{1+x^{2}} \end{aligned} \]

This is the Lorentzian (Cauchy) profile. A Gaussian \(e^{-x^{2}}\) is the transform of a Gaussian, not of \(e^{-|k|}\); \(e^{-|x|}\) simply copies the input; and \(1/(1+x)\) is not even an even function, while the transform of an even function must be even.

Discuss in Forum

Given that Fourier integral representation for the function

\[ f(x) = \begin{cases} 0 & \text{if } |x| \gt 1 \\ 1 & \text{if } |x| \lt 1 \end{cases} \]

is \( f(x)=\frac{2}{\pi}\int_0^\infty\frac{\cos\omega x\sin\omega}{\omega}d\omega \)

Which of the following options is correct ?

A.\( \int_0^\infty\frac{\cos\omega x\sin\omega}{\omega}d\omega \) = \( \pi/2 \) if \( 0 \le x \lt 1 \); \( \pi/4 \) if \( x = 1 \); 0 if \( x \gt 1 \)
B.\( \int_{0}^{\infty} \frac{\sin \omega}{\omega}\, d\omega = \pi \)
C.\( \int_{0}^{\infty} \frac{\sin \omega}{\omega}\, d\omega = \frac{\pi}{4} \)
D.\( \int_{0}^{\infty} \frac{\cos \omega x \sin \omega}{\omega}\, d\omega = 0 \)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The correct statement is the piecewise result: the integral equals \( \pi/2 \) for \( 0 \le x \lt 1 \), \( \pi/4 \) at \( x = 1 \) and 0 for \( x \gt 1 \).

Rearranging the given representation, the integral is \( \frac{\pi}{2} f(x) \) wherever \( f \) is continuous:

\[ \begin{aligned} &\int_0^\infty \frac{\cos\omega x \sin\omega}{\omega}\, d\omega \\ &= \frac{\pi}{2} f(x) \end{aligned} \]

So it is \( \pi/2 \) inside \( |x| \lt 1 \) and 0 outside. At the jump \( x = 1 \) a Fourier integral converges to the mean of the left and right limits, \( \frac{\pi}{2} \cdot \frac{1 + 0}{2} = \frac{\pi}{4} \).

Putting \( x = 0 \) gives the Dirichlet integral \( \int_{0}^{\infty} \frac{\sin \omega}{\omega} d\omega = \frac{\pi}{2} \), so the options giving \( \pi \) and \( \pi/4 \) are wrong, and the integral is 0 only for \( |x| \gt 1 \), not for every \( x \).

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Fourier series for the function

\[ f(\theta) = \begin{cases} +1 & 0 \lt \theta \lt \pi \\ -1 & \pi \lt \theta \lt 2\pi \end{cases} \]

there are :

A.Even cosine terms only
B.Both odd sine and cosine terms
C.Sine terms only
D.Odd sine terms only

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The series contains odd sine terms only.

The square wave is an odd function about \(\theta = 0\) (shifting the second half to \(-\pi \lt \theta \lt 0\) gives \(f(-\theta) = -f(\theta)\)), so every cosine coefficient and the constant term vanish. For the sine coefficients:

\[ \begin{aligned} b_{n} &= \frac{2}{\pi}\int_{0}^{\pi}\sin n\theta\,d\theta \\ &= \frac{2}{n\pi}(1 - \cos n\pi) \end{aligned} \]

This is \(4/n\pi\) for odd n and 0 for even n, giving \(f = \frac{4}{\pi}(\sin\theta + \frac{1}{3}\sin 3\theta + \dots)\). 'Sine terms only' is incomplete because the even harmonics are absent, and the cosine options fail the odd symmetry.

Discuss in Forum

Consider the function

\(f(x) = |x| \quad -1 \le x \le 1\)

Then the Fourier series for \(f\) is :

A.\(\frac{1}{2} - \frac{4}{\pi^{2}}\)\((\frac{\cos \pi x}{1^{2}} + \frac{\cos 3\pi x}{3^{2}} + \cdots)\)
B.\(\frac{1}{2} - \frac{4}{\pi^{2}}\)\((\frac{\sin \pi x}{1^{2}} + \frac{\sin 3\pi x}{3^{2}} + \cdots)\)
C.\(\frac{1}{2} - \frac{4}{\pi^{2}}\)\((\frac{\cos \pi x}{1^{2}} + \frac{\cos 3\pi x}{3^{2}} + \cdots)\) \(- \frac{4}{\pi^{2}}\)\((\frac{\sin \pi x}{1^{2}} + \frac{\sin 3\pi x}{3^{2}} + \cdots)\)
D.\(\frac{1}{2} + \frac{4}{\pi^{2}}\)\((\frac{\cos \pi x}{1^{2}} + \frac{\cos 3\pi x}{3^{2}} + \cdots)\) \(+ \frac{4}{\pi^{2}}\)\((\frac{\sin \pi x}{1^{2}} + \frac{\sin 3\pi x}{3^{2}} + \cdots)\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The Fourier series is \(\frac{1}{2} - \frac{4}{\pi^{2}}(\cos \pi x + \frac{1}{9}\cos 3\pi x + \cdots)\), the cosine series with odd harmonics only.

\(|x|\) is even, so it has a cosine series only (which rules out the options with sine terms). On [-1, 1], with period 2:

\[ \begin{aligned} a_{0} &= \int_{-1}^{1} |x|\,dx = 1 \\ a_{n} &= 2\int_{0}^{1} x\cos n\pi x\,dx \\ &= \frac{2((-1)^{n} - 1)}{n^{2}\pi^{2}} \end{aligned} \]

So \(a_{n} = -4/(n^{2}\pi^{2})\) for odd \(n\) and 0 for even \(n\), and the constant term is \(a_{0}/2 = 1/2\). The minus sign is right: at \(x = 0\) the series must give 0, and \(1/2 - (4/\pi^{2})(\pi^{2}/8) = 0\).

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Laplace transform of \(J_{0}(a\, t)\) where \(J_{0}\) denotes the Bessel function of zero order is :

A.\(\dfrac{1}{s^{2} + a^{2}}\)
B.\(\dfrac{1}{\sqrt{s^{2} + a^{2}}}\)
C.\(\dfrac{1}{s^{2} - a^{2}}\)
D.\(\dfrac{1}{\sqrt{s^{2} - a^{2}}}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The Laplace transform is \(\dfrac{1}{\sqrt{s^{2} + a^{2}}}\).

Using the series \(J_{0}(at) = \sum (-1)^{k}(at/2)^{2k}/(k!)^{2}\) and \(L\{t^{2k}\} = (2k)!/s^{2k+1}\), the transform sums to

\[ \begin{aligned} &\frac{1}{s}\left(1 + \frac{a^{2}}{s^{2}}\right)^{-1/2} \\ &= \frac{1}{\sqrt{s^{2} + a^{2}}} \end{aligned} \]

by the binomial series. \(1/(s^{2} + a^{2})\) is (up to a factor \(a\)) the transform of \(\sin at\), and the forms with \(s^{2} - a^{2}\) belong to the modified Bessel function \(I_{0}(at)\) and to \(\sinh at\).

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Fourier cosine transform of \(f(t) = e^{-t^{2}}\) is :

A.\(\dfrac{1}{\sqrt{2}}e^{-4s^{2}}\)
B.\(\dfrac{1}{\sqrt{2}}e^{-2s^{2}}\)
C.\(\dfrac{1}{\sqrt{2}}e^{-\frac{s^{2}}{2}}\)
D.\(\dfrac{1}{\sqrt{2}}e^{-\frac{s^{2}}{4}}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The Fourier cosine transform is \(\dfrac{1}{\sqrt{2}}e^{-s^{2}/4}\).

With the convention \(F_{c}(s) = \sqrt{2/\pi}\int_{0}^{\infty} f(t)\cos st\,dt\), and the standard integral

\[ \begin{aligned} \int_{0}^{\infty} e^{-t^{2}}\cos st\,dt &= \frac{\sqrt{\pi}}{2}e^{-s^{2}/4} \\ F_{c}(s) &= \sqrt{\frac{2}{\pi}}\cdot\frac{\sqrt{\pi}}{2}e^{-s^{2}/4} = \frac{1}{\sqrt{2}}e^{-s^{2}/4} \end{aligned} \]

The coefficient \(1/\sqrt{2}\) in every option shows this convention. The exponent must be \(-s^{2}/4\); the other exponents come from mis-scaling the Gaussian (\(-s^{2}/2\) is the transform of \(e^{-t^{2}/2}\)).

Discuss in Forum

If \(L\{f(t)\} = F(s)\) and \(L\{g(t)\} = G(s)\), then \(L^{-1}\{(F(s)G(s)\}\) is :

A.\(\int_{0}^{t} f(x)\, g(t - x)\, dx\)
B.\(\int_{0}^{t} f(x)\, g(x)\, dx\)
C.\(\int_{0}^{t} f(x)\, g(t + x)\, dx\)
D.\(\int_{0}^{t} f(x - t)\, g(x - t)\, dx\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

\(L^{-1}\{F(s)G(s)\}\) is \(\int_{0}^{t} f(x)\,g(t - x)\,dx\).

This is the convolution theorem: the Laplace transform of the convolution \((f * g)(t) = \int_{0}^{t} f(x)g(t - x)\,dx\) is the product \(F(s)G(s)\). It is proved by changing the order of integration in \(\int_{0}^{\infty} e^{-st}\int_{0}^{t} f(x)g(t - x)\,dx\,dt\) and substituting \(\tau = t - x\).

The inverse transform of a product is not the integral of the product \(f(x)g(x)\), and the arguments \(t + x\) or \(x - t\) in the other options do not give a convolution.

Discuss in Forum

10.

The complex Fourier transform of \(e^{-|x|}\) is :

A.\(\frac{2}{1 + n^{2}}\)
B.\(\frac{1}{1 + n^{2}}\)
C.\(\frac{4}{1 + n^{2}}\)
D.\(\frac{\pi}{1 + n^{2}}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The complex Fourier transform is \(\frac{2}{1 + n^{2}}\).

Using \(F(n) = \int_{-\infty}^{\infty} f(x)e^{inx}dx\) and splitting the range at zero, where \(|x|\) changes sign:

\[ \begin{aligned} F(n) &= \int_{-\infty}^{0} e^{(1 + in)x}dx \\ &\quad + \int_{0}^{\infty} e^{-(1 - in)x}dx \\ &= \frac{1}{1 + in} + \frac{1}{1 - in} \\ &= \frac{2}{1 + n^{2}} \end{aligned} \]

The paper uses the convention without the factor \(1/\sqrt{2\pi}\). With it, the result would be \(\sqrt{2/\pi}/(1 + n^{2})\), which is not offered. The other numerators come from dropping one of the two halves (1) or mishandling the constants (4 or \(\pi\)).

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