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The inverse change of variables is $x=v/(1+u)$, $y=uv/(1+u)$. Thus the triangular region becomesThe upper limit depends on $u$: the new region is not a rectangle. The absolute inverse Jacobian determinant is $x^2/(x+y)$, so the change of variables formula cancels the rational factor in the integrand:Although the original integrand is undefined at the vertex $x=y=0$, its nonnegative improper integral exists. One can apply the change of variables formula on $x\ge \epsilon$ and let $\epsilon \downarrow 0$ by the monotone convergence theorem; the transformed integrand has the finite integral just computed.

Use $u=1/x$, so $dx=-du/u^2$ and $\ln x=-\ln u$. Reversing the integration bounds gives, without evaluating either side,For $J={\int }_0^{\infty }\ln x/(1+x^2)dx$, the same change of variables givesso $\boxed{J=0}$. These are legitimate transformations of improper integrals: $|\ln x|$ is integrable near zero, and $(\ln x)/x^2$ is integrable at infinity. In particular $J$ is absolutely convergent, so no cancellation of divergent endpoints is being assumed.

Use the complex exponential representation of cosine. For real $\alpha$ and $\beta >0$, exponential decay makes the improper integral absolutely convergent, andFor the second integral, use differentiation under the integral sign with respect to $\beta$. The minus sign is essential: ${\partial }_{\beta }{e}^{-\beta x}=-x{e}^{-\beta x}$. HenceTo justify this step near any fixed ${\beta }_0>0$, restrict $\beta \ge {\beta }_0/2$. The absolute value of the differentiated integrand is at most $x{e}^{-{\beta }_{0}x/2}$, an integrable bound, so dominated convergence permits the differentiation. At $\alpha =0$ the answers reduce to $1/\beta$ and $1/{\beta }^2$, respectively.

For $x>0$ and $\alpha >1$, the preceding antiderivative givesThe integrand is nonnegative, so Tonelli's theorem permits reversing the two integrals without first assuming absolute integrability. Put $u=xy$ in the inner integral to obtain the hyperbolic-secant Frullani integral:Here ${\int }_0^{\infty }\mathrm{sech}u\tanh udu=[-\mathrm{sech}u{]}_0^{\infty }=1$. Near zero, cancellation makes the original integrand $O(x)$; at infinity it decays exponentially. The finite result therefore also establishes ordinary convergence of the improper integral.

With the change of variables $u=\tan x$, one has $dx=du/(1+u^2)$, $\sin x=u/\sqrt{1+u^2}$ and $\cos x=1/\sqrt{1+u^2}$. The transformed improper integral is integrable at its zero endpoint and becomesHence

For $x\in (n,n+1)$, strict decrease gives $f(x)<f(n)$, while for $x\in (n-1,n)$ it gives $f(x)>f(n)$. Integrating over these unit intervals provesSumming the rectangle bounds givesThe partial sums and the improper integral are nondecreasing. These bounds show that one is bounded exactly when the other is bounded, proving the integral test for convergence. In the convergent case, sum the same interval inequalities for $n=N+1,N+2,\dots$ to obtainFor $f(x)=1/(x\ln x)$, the antiderivative is $\ln (\ln x)$, which grows without bound, so the first series diverges. For $f(x)=1/[x(\ln x{)}^2]$, the antiderivative is $-1/\ln x$, whose tail integral is finite, so the second series converges, withBoth functions are positive and strictly decreasing on $[2,\infty )$, as required.

Set $t=\log (1/x)$, so $x=e^{-t}$ and $dx=-e^{-t}dt$. The improper integral becomesThe Gamma function recurrence givesThe exponential-transformed integrand also proves convergence at the logarithmically singular endpoint.

The Gaussian function $e^{-x^2}$ is even and $\sin x$ is odd, so their product is an odd function. Moreover $|e^{-x^2}\sin x|\le e^{-x^2}$, whose integral on the real line is finite. Thus the improper integral is absolutely convergent, not merely a symmetric principal value, and the contributions of its two half-lines cancel:

The expected value is the improper integral $\alpha {\int }_1^{\infty }{x}^{-\alpha }dx$. It converges exactly when $\alpha >1$, yieldingAt $\alpha =1$ this improper integral diverges logarithmically, and for $0<\alpha <1$ it diverges as a power of its upper endpoint. In both cases the nonnegative variable has infinite mean. This is the first Pareto moment threshold.

Index the stated geometric progression by $x_i=(1+ϵ{)}^i$, $i=0,1,\dots$. Then $\Delta x_i=ϵ(1+ϵ{)}^i$, and its left-endpoint contribution isHence the approximation is a geometric series withSince $ϵ>0$ and $\alpha >0$, the ratio satisfies $0<q<1$ and the series converges. The integrand is decreasing, so the left-endpoint sum is an upper bound for the improper integral.

For $m=2$, evaluate the improper integral directly:For $m\ge 3$, use integration by parts with $u=x^{m-2}$ and $dv=(1+x{)}^{-(m+n-2)}dx$. Then $v=-(1+x{)}^{-(m+n-3)}/(m+n-3)$ andThe boundary term vanishes: at zero $m-2>0$, while at infinity its power is $1-n<0$. The restriction $m\ge 3$ belongs to this recurrence step; the separate $m=2$ base case must not be replaced by a formal zero times a divergent $I_{1,n}$.

Use integration by parts in $I_{n+1}$, writing the integrand as $x^{2n+1}(x{e}^{-x^2})$. Since an antiderivative of $x{e}^{-x^2}$ is $-e^{-x^2}/2$,The boundary term vanishes because the Gaussian exponential function decays faster than any fixed power. These improper integrals are absolutely convergent, justifying the limiting integration by parts.

Starting from the permitted Gaussian integral $I_0=\sqrt{\pi }$, the Gaussian moment recurrence gives $I_1=\sqrt{\pi }/2$ and thenMore generally, iteration gives $I_n=(2n-1)!!\sqrt{\pi }/2^n$, using the double factorial with $(-1)!!=1$.

The function $g(x)=1/(x\ln x)$ is positive and decreasing for $x\ge 2$, since $g^{\prime }(x)=-(\ln x+1)/(x^2{\ln }^{2}x)<0$. The integral test for convergence therefore applies. Its comparison improper integral isConsequently $\sum _{n=2}^{\infty }1/(n\ln n)$ diverges.

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.

For $n>1$, integration by parts with $u=x^{n-1}$ and $dv=e^{-x}dx$ yieldsThe boundary term vanishes at zero because $n-1>0$, and at infinity because exponential decay dominates every fixed power. The initial improper integral is $I_1={\int }_0^{\infty }{e}^{-x}dx=1$. Iterating the recurrence relation therefore givesThis is the positive-integer case of the Gamma function identity $\Gamma (n)=(n-1)!$.

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

Write $q(x)=1/[x(1+(\ln x{)}^2)]$. The vector field is undefined at $x=0$ because $q$ diverges there, but this component is tangential to that face. The face fluxes have finite improper integral limits. In particular, the substitution $s=\ln x$ givesUsing the outward unit normal on each face, the flux contributions areHenceFor a precise justification at the singular face, first use the truncated cube $\epsilon \le x\le 1$. Its opposite $y$-face integrals cancel, the $x$-faces contribute $e^{-1}-e^{-\epsilon }$, and the upper $z$-face contributes $1-\epsilon$. The limit as $\epsilon \downarrow 0$ is $e^{-1}$. On each truncated cube the divergence theorem gives the same answer, since $\nabla \cdot \mathbf{F}=1-e^{-x}$. Thus the result is a flux of a vector field with an integrable tangential singularity, rather than an application of the classical smooth-field theorem directly at $x=0$.

Let $I$ be the positive Gaussian integral. Its improper integral converges because $e^{-x^2}\le e^{-|x|}$ for $|x|\ge 1$. Multiplying two copies and using the polar coordinates area element givesNonnegativity justifies the product double integral and the change of integration order. Since $I>0$,

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

The product in the integrand simplifies to $\sin x{\cos }^{3}x$ below its upper endpoint. With the integration by substitution $t=\cos x$, the limits become $c$ and zero and $dt=-\sin xdx$. ThereforeAlthough the tangent function itself is undefined at the upper endpoint, the product has a zero limit there, so its continuous extension, or its improper integral, gives this same result.

To obtain $J$ directly from a Riemann sum, divide $[0,c]$ into $n$ equal pieces of width $c/n$ and use right endpoints $t_k=kc/n$. The sum isThe integrand is continuous, so the limit of these Riemann sums is its definite integral, namely $c^4/4$, agreeing with the antiderivative calculation.