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The integrand for $I_1$ is an odd function with absolute convergence, so $I_1=0$. For $I_2$, the recurrence gives $I_2=I_0/2$, where the supplied Gaussian integral is $I_0=\sqrt{\pi }$. HenceThe zero first moment is specific to the symmetric full-line domain.

The argument $3x^3+2x$ is odd, so its squared sine is even. Replacing $x$ by $-x$ interchanges the numerator and denominator inside the logarithm, changing that factor's sign. Thus the full integrand is an odd function. It is continuous on the symmetric interval because $|x|\le 1/\pi <1$. Therefore

The required property is odd symmetry, $\boxed{f(-x)=-f(x)}$, up to irrelevant changes on a set of measure zero when discussing Fourier coefficients. An odd function has zero mean, and its products with cosine are odd, so every cosine coefficient vanishes. Its sine coefficients may remain nonzero.

For pointwise representation, the Fourier series takes the average of the one-sided limits at a jump. In particular a pure Fourier sine series has value zero at the origin and at the endpoints; arbitrary assigned values there need not equal the series sum.

Use the periodic extension of $f$ when the translated argument falls outside the original interval. With $k_n=2\pi n/L$, the sine subtraction identity givesThis translation of a Fourier series generally introduces cosine terms: a translated odd function need not be odd about the original origin. Equality is understood wherever the original Fourier series represents its periodic extension, with midpoint values at jumps.

The same integration by substitution gives the expected valueThe remaining integrand is an odd function. Its integral is absolutely convergent, since ${\mathrm{sech}}^{2}t\le 4e^{-2|t|}$ and $|t|e^{-2|t|}$ is integrable. Thus cancellation under symmetry in integration is legitimate, rather than merely a principal-value calculation. ConsequentlyEquivalently, the logistic distribution is symmetric about $\mu$, and its exponentially decaying tails ensure that its expected value exists.

Each of the three surface pieces is invariant under $y\mapsto -y$, while $y^3$ is an odd function under that reflection. Its surface integral therefore vanishes on each piece separately. Only $z$ contributes.

On the cylindrical side, use cylindrical coordinates $x=a\cos \varphi$, $y=a\sin \varphi$, with area element $dS=ad\varphi dz$:The upper disc has constant $z=b$ and area $\pi a^2$, giving $I_{\mathrm{top}}=\pi a^{2}b$. On the hemispherical indentation, use spherical coordinates with $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\varphi$:Adding the three contributions yieldsThis is a scalar surface integral, so the indentation contributes positively through $dS$; an inward or outward orientation does not change its sign.

On the open interval, $g=f^{\prime }$. It is an odd function there, so its constant and cosine coefficients vanish; assigned values at the endpoints do not affect these integrals. For its sine coefficient, integration by parts givesConsequentlyThe endpoint qualification matters. The periodic $g$ has a jump discontinuity at every odd integer, with one-sided limits $\pm \sinh 1$. By the Dirichlet-Jordan convergence theorem, the Fourier series converges to their average, zero, there, not to the assigned value $g(1)=\sinh 1$. The integration by parts argument also avoids assuming unrestricted termwise differentiation of the first series.

For $n=0$, the change of variables $s=\sqrt{a}x$ reduces the integral to the given Gaussian integral:For $n=1$, the integrand is an integrable odd function, so its contributions from the two half-lines cancel:The condition $a>0$ ensures convergence in both cases.

Since the probability density function is an even function, $xf(x)$ is an odd function. Its integral over the symmetric support vanishes, so

For $L=1$, the even function $h(x)=x^2$ has $b_n=0$, because $x^2\sin (n\pi x)$ is an odd function. Its constant coefficient is $a_0={\int }_{-1}^1{x}^{2}dx=2/3$. The remaining Fourier coefficients areTwo applications of integration by parts give the antiderivative, with $k=n\pi$,Its endpoint difference is $2(-1{)}^n/(n\pi {)}^2$, so $a_n=4(-1{)}^n/(n^2{\pi }^2)$. ThereforeThe periodic extension is continuous and piecewise smooth, so pointwise convergence of a piecewise smooth Fourier series gives equality, including at the endpoints.

Reflection in the $y$-axis preserves the disk but changes the sign of the integrand $xy$. The double integral is therefore its own negative and must beEquivalently, for each fixed $y$ the integrand is an odd function of $x$ on a symmetric interval.

Since $q$ is an even function, $tq(t)$ is an odd function and its integral over the symmetric interval vanishes. Thus $\mathbb{E}[T]=0$. Its variance isConsequentlyThe standard deviation is approximately $0.40825$ minute, or $24.49$ seconds.

The function is an odd function, so $a_0=a_n=0$. Here $L=1$, and the Fourier coefficients areWrite $k=n\pi$. One integration by parts has zero endpoint terms because $x(1-x)$ vanishes at both endpoints:A second integration by parts givesTherefore $b_n=4(1-(-1{)}^n)/(n^3{\pi }^3)$, zero for even $n$ and $8/(n^3{\pi }^3)$ for odd $n$. The Fourier series isThe coefficients are absolutely summable, and the periodic function is continuous even at the joins $x=\pm 1$, so the displayed Fourier series represents it everywhere.

The factor $x^2$ is an even function and $\sin x$ is an odd function, so their product is an odd function. Its definite integral over the symmetric interval vanishes:Indeed the substitution rule $x=-t$ changes the integral over $[-\pi /2,0]$ into the negative of the integral over $[0,\pi /2]$.

A probability density function must have total integral one. Put $h=\pi /6$. The term $ax$ is an odd function, soThus $\boxed{b=1}$.

For $h=\pi /6$, the expected value isbecause $x\cos x$ is an odd function. Comparing with the specified expected value gives $\boxed{a=1}$, which lies in the permitted parameter range.

Here $a=0$, so the probability density function is the even function $\cos x$ on $[-h,h]$, where $h=\pi /6$. Hence $\mathbb{E}[X]=0$ by oddness, and the variance is just $\mathbb{E}[X^2]$. Two applications of integration by parts giveThus

The function is even, so each product with a sine is an odd function and all the Fourier coefficients $b_n$ vanish. Its constant coefficient isFor $n\ge 1$, part (a) gives the remaining Fourier coefficients:Consequently the Fourier cosine series isThe periodic extension is continuous, including at $\pm \pi$, and piecewise continuously differentiable. Thus the Fourier series equals $f$ everywhere. Its coefficients are also bounded in magnitude by a constant times $1/n^2$, which gives absolute convergence.

An antiderivative of the initial velocity is $-2c{e}^{-s^2/2}$. Thus D'Alembert's formula givesThis is a positive Gaussian function moving right and a negative Gaussian function moving left. Their cancellation at $t=0$ gives the zero initial displacement, while direct differentiation gives ${\psi }_t(x,0)=2cx{e}^{-x^2/2}$.

For every $t>0$, the profile is an odd function of $x$, vanishes at $x=0$, is positive for $x>0$ and negative for $x<0$, and tends to zero at both spatial infinities. The signs follow from $(x+ct{)}^2-(x-ct{)}^2=4xct$. The positive and negative lobes move farther apart as $t$ increases; for large separation they approach the individual unit-height Gaussian functions centred at $x=\pm ct$. At finite separation the actual extrema lie slightly beyond these centres: writing $a=ct$, the positive global maximum occurs at the unique $x>a$ satisfying $(x-a)/(x+a)=e^{-2ax}$.

For the odd extension, the cosine integral vanishes and the sine integral doubles. Integration by parts givesThe value at $k=0$ is zero by the odd function symmetry, or by the Maclaurin series of the sine. The transform is purely imaginary and odd.