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The integral converges absolutely: near zero it behaves like $\log x$, and at infinity like $(\log x)/x^2$. Let its value be $J$. The integration by substitution $x=1/t$ givesThereforeAbsolute convergence makes this cancellation between the two ends legitimate.

Put $R=\sqrt{8}$ and parametrize the circle counterclockwise, as viewed from positive $z$, by $\mathbf{r}(\varphi )=(R\cos \varphi ,R\sin \varphi ,3)$ for $0\le \varphi \le 2\pi$. The restricted field isThe PDF does not specify the differential in this line integral. An oriented vector line integral weighted by a scalar field uses $d\mathbf{r}=(-R\sin \varphi ,R\cos \varphi ,0)d\varphi$. Its two nonzero components areAll other terms vanish by sine-cosine symmetry. Thus this interpretation gives

If the intended convention is the scalar arc length integral, then $ds=Rd\varphi$ andThat scalar integral is independent of traversal direction. Giving both values retains the actual distinction left unspecified by the wording.

For $m>-1$, integrate the base power to obtain $J(m)={\int }_0^1{x}^{m}dx=1/(m+1)$. Since ${\partial }_m^n{x}^m=x^m(\log x{)}^n$, differentiating $n$ times givesHere $n$ is a nonnegative integer. For justification, keep $m$ in a small closed interval bounded below by $m_0>-1$. The differentiated integrands are dominated by $x^{m_0}|\log x{|}^n$, whose integral becomes ${\int }_0^{\infty }{e}^{-(m_0+1)t}{t}^{n}dt$ under $x=e^{-t}$ and is finite. The dominated convergence theorem establishes the needed differentiation under the integral sign and the logarithmic moments of a power on the unit interval.

Let $D=3x^2+2x+2$. Splitting the numerator into a derivative term and a constant gives $x-1=D^{\prime }/6-4/3$. ThusComplete the square as $D=((3x+1{)}^2+5)/3$ and substitute $u=3x+1$. The remaining integral is $\int du/(u^2+5)=\arctan (u/\sqrt{5})/\sqrt{5}$. HenceThe quadratic denominator is positive for all real $x$, so no logarithmic absolute value or real-domain exclusion is needed.

Integrate the normal component on each face directly, with outward normals. Each unlisted pair of coordinates ranges independently from zero to one:For example, on $x=1$ the integral is ${\int }_0^1{\int }_0^{1}2(y+z)dydz=2$. Adding all six contributions givesThis calculation uses the six explicit surface integrals, independently of the divergence theorem.

Assume real square-integrable functions on the same integration domain. Put $A=\int f^2$, $B=\int g^2$, and $C=\int fg$. If $B=0$, then $g=0$ almost everywhere, so $C=0$ and the claim follows. If $B>0$, nonnegativity of a square givesMultiplying by $B$ proves the integral form of the Cauchy-Schwarz inequality:For nonzero functions, equality holds exactly when they are proportional almost everywhere. The zero-function cases also give equality.

The integrand and both limits depend on $x$, so use the Leibniz integral rule. The upper and lower endpoint contributions and the interior derivative giveFor $x\ne 0$, an antiderivative in $t$ is $t\sin (xt)/x+\cos (xt)/x^2$. Evaluating it and collecting terms givesAt $x=0$ the original Leibniz expression has zero endpoint terms and an integral over a zero-length interval, giving $\boxed{F^{\prime }(0)=0}$. The displayed quotient has a removable singularity and the same limiting value.

The parameter derivative of the integrand is $\cos (a\log x)$, whose absolute value is bounded by one on $0<x<1$. This integrable bound justifies differentiation under the integral sign, givingThe original integral also exists: $|\sin (a\log x)/\log x|\le |a|$, and its apparent singularity at $x=1$ is removable. Set $x=e^{-t}$ to evaluate the derivative:ConsequentlySince $I(0)=0$, integration in the parameter fixes the constant and gives the logarithmic sine integral identity

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.

For $s>0$, integrate the probability density function to obtain the cumulative distribution functionIf $\mu >0$, the requested probability is $F_X(\mu )-F_X(0)$. Since the hyperbolic tangent is odd, this givesThe second case matters because the question places no sign restriction on $\mu$: the event is empty when its upper endpoint is nonpositive. Using an oriented integral from $0$ to a negative $\mu$ would give a negative number, not a probability.

The function $g(x)=1/(x{\ln }^{2}x)$ is likewise positive and decreasing on $[2,\infty )$, with $g^{\prime }(x)=-(\ln x+2)/(x^2{\ln }^{3}x)<0$. NowThe integral test for convergence shows that $\sum _{n=2}^{\infty }1/(n{\ln }^{2}n)$ converges. This comparison integral determines convergence, not the numerical value of the series itself.

Reverse the order in the nonnegative double integral, using $0\le y<\infty$ and $y\le x\le 2y$. For $y>0$, an antiderivative in $x$ is $-(x/y+1/y^2)e^{-xy}$. ThusLet $J$ denote the requested improper integral. Its numerator is $y^2+O(y^4)$ near zero, so its apparent singularity is removable; it also decays at infinity. We may therefore separate the preceding expression into the convergent integralsIt follows thatThe two terms divided separately by $x^2$ would each have a divergent integral at zero; the cancellation in their difference is essential.

The stated derivative formula needs a domain qualification. The original PDF calls $\alpha$ real, but the integral is even in $\alpha$; its derivative therefore cannot equal the printed positive expression on both sides of zero.

First take $\alpha >0$. The improper integral converges, with an $O(x^2)$ integrand near zero and $O(\ln x/x^2)$ decay at infinity. On compact positive parameter intervals, an integrable bound also justifies differentiation under the integral sign:For $\alpha \ne 1$, the partial fraction decomposition isIntegrating the rational functions givesAt $\alpha =1$, the same result follows by continuity, or directly from $2{\int }_0^{\infty }{x}^2/(1+x^2{)}^{2}dx=\pi /2$.

Integrating gives $E(\alpha )=\pi \ln (1+\alpha )+C$ on the positive half-line. To fix $C$, take $\alpha \to 0^{+}$: for $0\le \alpha \le 1$ the integrand is bounded by $\ln (1+x^2)/(1+x^2)$, an integrable function. The dominated convergence theorem gives $E(\alpha )\to E(0)=0$, so $C=0$. Evenness then extends the logarithmic parameter integral with Cauchy weight to its full stated domain:For negative $\alpha$, the derivative is $-\pi /(1-\alpha )$. At zero the right and left derivatives are $\pi$ and $-\pi$, so $E$ is not differentiable there. The pointwise derivative of the integrand at zero is zero, but interchanging it with the infinite integral at that parameter is invalid. Finally, substituting $\alpha =1$ gives the requested special value

For fixed $r<1$, $\sum r^n\cos (n\theta )$ converges absolutely and uniformly. Thus for integrable $f$, one may integrate the kernel series term by term. The cosine addition formula givesSubstitute this into the defining integral. The constant kernel term gives $\frac{1}{2\pi }\int f=a_0/2$; the remaining terms givewhereThese are the same Fourier coefficients as in part (a). The kernel $2K$ is the Poisson kernel for the unit disk; the factors $r^n$ suppress the higher angular frequencies inside the disk.

The boundary data are odd except at isolated endpoint choices, which do not change integrals. Thus every cosine Fourier coefficient, including $a_0$, vanishes. For the sine Fourier coefficients,Consequently the extension isIt converges absolutely for every $r<1$; its limiting boundary values at jumps are the midpoint of the two one-sided limits.

At a fixed radius, Parseval's identity applied to the preceding Fourier series givesIn polar coordinates, $dA=rdrd\theta$. Integrating over radius and interchanging the nonnegative sum and integral givesThe interchange follows from the monotone convergence theorem. This is the special case of the area norm of the Poisson extension in the unit disk for the odd step boundary data.

Use the law of total probability to average the conditional interval probability over $Y$:The integrands are nonnegative, so Tonelli's theorem permits interchanging the integrals. ThusReading off the probability density gives the convolution of probability densities:

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 the period-two Fourier series of a square-integrable function $w$, Parseval's identity statesFor $w=f-g$, write $s=\sinh 1$. Its coefficients are $a_0=2s$, $a_n=2(-1{)}^{n}s/(1+n^2{\pi }^2)$ and $b_n=n\pi a_n$, because $g$ is subtracted. ThereforeThe sum was obtained in part (a). As a separate check, $f-g=e^{-x}$ in the interval, whose square has integral ${\int }_{-1}^1{e}^{-2x}dx=\sinh 2$. Endpoint jumps have no effect on this integral.

The divergence isFor the tetrahedral domain, convenient integration bounds areConsequently its volume integral isPutting $h=2(1-x-y)$, the inner integral is $(2y+6)h+h^2=(16-4x)(1-x-y)$. ThereforeThusThere is an independent centroid check. The vertices are $(0,0,0),(1,0,0),(0,1,0),(0,0,2)$, giving volume $1/3$ and centroid $(1/4,1/4,1/2)$. An affine function averages to its value at the centroid, so $I=(1/3)(6+2/4+2/2)=5/2$. This is the integral of an affine function over a simplex.