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Fubini's theorem states that an absolutely integrable measurable function on a product of two $\sigma$-finite measure spaces has integrable sections almost everywhere, the corresponding iterated integrals exist, and both orders equal its product integral. In particular, for $f\in L^1({\mathbb{R}}^2)$,To establish integrability of the nonnegative function here, first use Tonelli's theorem: for a nonnegative measurable function the product integral equals either iterated integral with infinity allowed. This avoids assuming the conclusion needed for Fubini's theorem.

If $\alpha \ge -1$, ${\int }_0^{\infty }(1+x+y{)}^{\alpha }dx=\infty$ for every fixed $y$. If $\alpha <-1$, it isThe outer integral is finite exactly when $\alpha +1<-1$, that is, $\alpha <-2$. In that caseFor $-2\le \alpha <-1$ the outer integral diverges, including logarithmic divergence at $\alpha =-2$. Nonnegativity means finiteness of this integral is exactly absolute integrability; after proving it, Fubini's theorem allows either order.

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus isIt changes width and height with time, so this is not a stationary state. Its integral is the constant $\sqrt{\pi \hslash /m}$; multiplying the wavefunction by $(m/(\pi \hslash ){)}^{1/4}$ normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance $\hslash (1+t^2)/(2m)$. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

Use $|\sin (\alpha x)|\le |\alpha |x$ for $x\ge 0$. For the source's functions,The last integral follows by the substitution $u=nx$ and integration by parts. Since ${\sum }_n{n}^{-2}$ converges, (ii) permits summing the integrals and identifies an integrable almost-everywhere sum.

For every $x>0$, the geometric series givesThe functions vanish on $x\le 0$, and the single endpoint does not change the integral. Use the supplied integral of $f_n$ to obtainThe estimates include negative $\alpha$ through its absolute value, and the case $\alpha =0$ gives zero on both sides.

Put $G_N={\sum }_{n=1}^N|f_n|$. These are nonnegative measurable functions increasing to $G={\sum }_{n=1}^{\infty }|f_n|$, possibly initially with value infinity. Monotone convergence theorem givesAn infinite value on a set of positive measure would force an infinite integral. Thus $G$ is finite almost everywhere and, after setting it to zero on its null exceptional set if desired, is integrable.

Where $G$ is finite, the real series ${\sum }_n{f}_n$ is absolutely convergent. Define its sum to be $h$ there and zero on the exceptional null set. Measurability follows as an almost-everywhere limit of partial sums, and $|h|\le G$ almost everywhere gives integrability. Every partial sum is dominated by $G$, so dominated convergence theorem yieldsIn fact the convergence is also in the integrable-function norm:

The monotone convergence theorem says that if $0\le f_n\uparrow f$ almost everywhere and the functions are measurable on a measure space, thenwith infinite integrals permitted. The limit is measurable after the usual modification on any exceptional null set.

The dominated convergence theorem says that if measurable functions $f_n$ converge almost everywhere to $f$ and satisfy $|f_n|\le g$ almost everywhere for an integrable nonnegative function $g$, then $f$ is integrable andThe theorem applies equally to real or complex functions. In particular domination must be by one fixed Lebesgue integrable function, rather than by a different bound for each $n$.

For fixed real $t$, multiplication by $e^{itx}$ does not change the absolute value. The product integrability proved in (ii) therefore justifies Fubini theorem in the following calculation. Set $u=x-y$ in the inner integral:ThusThe positive exponential convention in the source is used throughout; no normalization factor is introduced under that convention.

Choose measurable versions of the two Lebesgue integrable functions and consider $F(x,y)=g(x-y)h(y)$. Translation and the corresponding linear change of variables preserve Lebesgue measurability; one can first use Borel versions and then disregard the resulting null-set changes. Tonelli's theorem and translation invariance giveFubini theorem therefore shows that the section in $y$ is integrable for almost every $x$. Define $(g\ast h)(x)$ by that section integral where it exists and by zero on the exceptional null set. Fubini theorem also makes this a measurable Lebesgue integrable function, withIts almost-everywhere defining formula is precisely the desired convolution. This construction explains why an everywhere-existing pointwise integral is not required.

Fubini's theorem for the product of two sigma-finite measure spaces states that an integrable measurable function on the product has integrable sections almost everywhere, the section integrals define Lebesgue integrable functions, and both iterated integrals equal the product integral. The integrability hypothesis is absolute integrability. For nonnegative measurable functions, Tonelli's theorem gives equality of the possibly infinite product and iterated integrals without first assuming finiteness.

The displayed function is not Lebesgue integrable. Its value at the origin can be assigned arbitrarily, since a point is a null set. On $0<x<1$ and $0<y<x/2$,Consequently Tonelli's theorem gives the divergent lower boundThe signed iterated integrals illustrate why the hypothesis matters. For $x>0$, $f(x,y)={\partial }_y[y/(x+y{)}^2]$, so integration first in $y$ and then in $x$ gives $1/2$. Integrating first in $x$ gives $-1/2$, using $f(x,y)=-{\partial }_x[x/(x+y{)}^2]$. Exceptional boundary sections have measure zero. Thus finite iterated values do not establish product integrability.

For the upper function graph, the vertical cross-section of the enclosed region has length $y(x)$, so its area is $J[y]={\int }_0^{2a}y(x)dx$. A maximizing nonnegative arch therefore uses $F=y$ and $G=\sqrt{1+y^{\prime 2}}$ in the normal Euler-Lagrange equation. It becomesIntegrating once gives $y^{\prime }/\sqrt{1+y^{\prime 2}}=(c-x)/\lambda$. The concave upper branch has $\lambda >0$ and integrates to $y=\sqrt{{\lambda }^2-(x-c{)}^2}+d$. The equal zero endpoint heights force $c=a$ and $d=-\sqrt{{\lambda }^2-a^2}$. HenceIt is a circular arc. To find its parameter, substitute into the length integral: $\sqrt{1+y^{\prime 2}}=\lambda /\sqrt{{\lambda }^2-(x-a{)}^2}$, givingThis scalar equation determines $\lambda$ numerically when needed; no explicit solution is required.

The arch is a global maximum in its smooth function graph regime. For $\lambda >0$, the integrand $y-\lambda \sqrt{1+y^{\prime 2}}$ is concave and strictly concave in its slope. Its supporting-tangent inequality, integrated against any competing function graph with the same endpoint values, has first-order term zero by the Euler-Lagrange equation. Thus $J[y]-\lambda I[y]\le J[y_{\ast }]-\lambda I[y_{\ast }]$. On the fixed-length class this is $J[y]\le J[y_{\ast }]$; strict slope concavity also gives uniqueness.

The source leaves the length range implicit. A twice continuously differentiable function graph of this form needs $\lambda >a$, giving $2a<l<\pi a$. At $l=2a$ the straight function graph is the degenerate solution. The semicircle limit $l=\pi a$ has vertical endpoint tangents and is outside that smooth function graph class. For $l>\pi a$, no arc in the claimed smooth function graph family exists, so the stated formula cannot describe a twice continuously differentiable maximizing function graph under those unrestricted data. Allowing overhanging curves changes the admissible class.

Set $u=r\psi$, choose $\alpha >0$ for a negative bound-state energy, and write $\kappa =2mK/{\hslash }^2$. Multiplying the radial equation by $r$ givesFor $u=e^{-\alpha r}f$, its second derivative is $e^{-\alpha r}(f^{\prime \prime }-2\alpha f^{\prime }+{\alpha }^{2}f)$. Substitution cancels the two ${\alpha }^2$ terms and givesFor the physical hydrogen problem, the angular eigenvalue has $\ell$ a nonnegative integer. Regularity at the origin selects the Frobenius exponent $u\sim r^{\ell +1}$, rather than the singular branch. PutThe equation for the series factor $v=f/r^{\ell +1}$ is $r{v}^{\prime \prime }+(2\ell +2-2\alpha r)v^{\prime }+(\kappa -2\alpha (\ell +1))v=0$, and coefficient comparison yieldsThe regular confluent-hypergeometric solution generically grows like $e^{2\alpha r}$ in this series factor; after the extracted $e^{-\alpha r}$ it is not normalizable. A bound state requires termination at some degree $q\ge 0$, namely $2\alpha (q+\ell +1)=\kappa$. Thus $n=q+\ell +1$ is an integer greater than $\ell$, andRegularity, normalizability and the integer angular index are the physical conditions selecting this spectrum; the differential equation without those conditions would admit many other solutions.

For $n=1$, necessarily $\ell =0$ and $q=0$. With $a_0={\hslash }^2/(mK)$,A normalized spatial ground state is ${\Psi }_1=(\pi a_0^3{)}^{-1/2}{e}^{-r/a_0}$. In spherical coordinates the radial probability has weight $r^{2}dr$. Constants and the angular integral cancel in the tail ratio. The supplied integral with $\lambda =2/a_0$ gives

Use the nonnegative-part convention $h=h^{+}-h^{-}$, where $h^{+}=\max (h,0)$ and $h^{-}=\max (-h,0)$. The zero-mean assumption givesFor $k\ge 1$, write the integral over the $k$th period as $I_k=P_k-N_k$, whereBoth $P_k$ and $N_k$ lie in the interval $[A/((k+1)\pi ),A/(k\pi )]$. Their difference therefore satisfies the stronger estimateConsequently ${\sum }_{k\ge 1}{I}_k$ converges absolutely as a series of numbers, proving the requested limit of integrals over whole periods. This does not assert absolute integrability of the original function: the cancellation took place inside each period.

Indeed, setting $B={\int }_0^{\pi }|h(t)|dt$, we haveThe harmonic series diverges, so $g$ is Lebesgue integrable only if $B=0$. Conversely $B=0$ means $h=0$ almost everywhere on one period, and hence on all periods, so $g=0$ almost everywhere. ThereforeThis is an example of integrating a mean-zero periodic function divided by x.

The partial sums $s_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase to an extended-valued function $s$. By the monotone convergence theorem,A nonnegative function with finite Lebesgue integral cannot equal $+\infty$ on a set of positive measure: otherwise its integral would exceed $M$ times that measure for every $M>0$. Thus the series converges to a finite value almost everywhere. Redefine it to be zero on the exceptional null set if a finite-valued representative is desired. The resulting function is Lebesgue integrable, with

Put $x=e^{-t}$. Thenwhere the second substitution is $u=(k+1)t$ and the final integral is the given gamma function value. The formula also holds for $k=0$, with value $1$.

For $0<x<1$, the exponential function expansion givesEvery term is nonnegative. By the monotone convergence theorem, integration can be interchanged with summation, and henceThis identity is one form of Sophomore's dream.

Expand $x^x=e^{x\log x}$. The absolute values of all terms have total integralThe series converges, for instance because $j^{-j}\le 2^{-j}$ for $j\ge 2$. Part (b) applied to the absolute values, followed by subtraction of positive and negative parts, therefore justifies termwise integration. It gives the other Sophomore's dream identity:Equivalently, the partial sums are dominated by $e^{|x\log x|}\le e^{1/e}$, so the dominated convergence theorem gives the same justification.

The stationary Schrödinger equation isThe Laplacian represents the kinetic-energy operator. For a spherically symmetric function, $\Delta \psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$. The attractive Coulomb potential of the nucleus is $V=-q^2/(4\pi {ϵ}_{0}r)$; $E$ is the energy eigenvalue. Multiplication by $-2m/{\hslash }^2$ produces the displayed radial equation with Bohr radiusFor a normalizable bound state, write ${\kappa }^2=-2mE/{\hslash }^2$ and take $\lambda >0$. Substitution of $\psi =(A+Br)e^{-\lambda r}$ ingives, after division by the exponential, the coefficientsIf $B=0$ and $A\ne 0$, these force $\lambda =1/a$ and $\kappa =\lambda$. If $B\ne 0$, the first coefficient gives ${\kappa }^2={\lambda }^2$, the second gives $\lambda =1/(2a)$, and the third gives $B=-A/(2a)$. Up to nonzero scalar factors, the two radial bound states and their energies are consequentlyThey are linearly independent as functions and correspond to different energy eigenvalues. Strictly, they are not two solutions of one fixed-energy differential equation: each is a solution at its own energy.

For the higher-energy state, the spherical Born rule includes the radial factor $4\pi r^2$. Its normalization constant and the angular factor cancel in the ratioThe given integral formula makes the denominator $8a^5$ and the numerator $8a^5-21a^5/e$. Thus

The function is nonnegative, measurable, and even. Direct integration givesThus it is integrable, with integral $\boxed{2}$.

On finite intervals the function is continuous, butConsequently its nonnegative integral on the half-line is infinite: it is not integrable. The divergence occurs at infinity, not at the origin.

The monotone convergence theorem states that if measurable functions $0\le f_n\uparrow f$ pointwise on a measure space, then $\int f_n\uparrow \int f$, allowing the limit to be infinite.

Extend each restriction to $I_n$ by zero on $I\setminus I_n$. The functions $f{1}_{I_n}$ are nonnegative and increase pointwise to $f$ because the intervals are increasing and exhaust $I$. HenceFor nonnegative $f$, integrability means finiteness of this integral, soNo boundedness or closedness of the intervals is required.

For a continuously differentiable diffeomorphism $\varphi :J\to I$ between intervals, the integration by substitution statesIt holds as an identity element of extended integrals for nonnegative measurable $h$. For signed or complex-valued measurable $h$, integrability on one side is equivalent to integrability after substitution, and the identity element then holds. The absolute derivative accounts for a decreasing substitution; on oriented finite intervals the equivalent formula uses ${\varphi }^{\prime }$ and oriented endpoints.

Put $C={\int }_0^{\infty }(1+x^4{)}^{-1}dx=\pi /(2\sqrt{2})$. The map $x=y^{-1/4}$ is a decreasing diffeomorphism of the positive half-line onto itself andApplying the nonnegative version before assuming finiteness proves both existence and value:Tonelli's theorem permits either iterated integral of a nonnegative measurable function on a product of sigma-finite measure spaces and identifies both with its extended product integral. Fubini's theorem states that for an absolutely integrable measurable function on such a product, almost every section is integrable, the section-integral functions are integrable, and the two iterated integrals equal the product integral.

The given two-variable function is continuous and positive on the positive quadrant. For fixed $y>0$, substituting $t=x{y}^{1/4}$ givesTonelli now givesThis proves absolute integrability, so Fubini is applicable.

For $x\ne 1$, integrate the displayed partial fraction identity element in $y$ up to $R$:At $x=1$ the section integral is ${\int }_0^{\infty }(1+y{)}^{-2}dy=1$, agreeing with the continuous limit of this expression. Therefore Fubini identifies the remaining integral and givesThe apparent singularity at one is removable, with limiting integrand value $1/4$. Positivity and the finite product integral already establish convergence at both ends.

Take a smooth variation $u+\epsilon v$ with $v=0$ on the boundary. Differentiating under the integral gives the first variationIntegration by parts removes the derivatives of $v$, and the boundary contribution vanishes. The fundamental lemma of the calculus of variations therefore givesFor a regular constraint $J[u]=c$, the Lagrange multiplier rule says that an extremal satisfies $\delta I=\lambda \delta J$ for a constant $\lambda$. Equivalently, apply the preceding Euler–Lagrange equation to $F-\lambda G$. Regularity means that $\delta J$ is not identically zero as a linear functional on the admissible variations. For the quadratic constraint below this holds at every nonzero admissible $u$; a zero constraint leaves only $u=0$.

In polar coordinates the constrained integrand isIts Euler–Lagrange equation is $-2\lambda ru-{\partial }_r(2r{u}_r)-{\partial }_{\theta }(2u_{\theta }/r)=0$, orThe field is single-valued and periodic in $\theta$. Substitution $u=f(r)\cos (n\theta )$ with $n\in \{0,1,2,\dots \}$ givesThe centre is part of the disk, so one must also select the regular radial solution: for $n\ge 1$, smoothness gives $f(r)=O(r^n)$ and the regular derivative behaviour; for $n=0$, $f$ is bounded and $f^{\prime }(0)=0$. A singular solution at zero does not define an admissible smooth field on the disk.

For fixed $n$, let $c_n={\int }_0^{2\pi }{\cos }^2(n\theta )d\theta$, so $c_0=2\pi$ and $c_n=\pi$ for $n\ge 1$. The angular derivative term has integral $n^2{c}_n$, including zero when $n=0$. ConsequentlyThe one-dimensional Euler–Lagrange equation for $\hat{I}-\lambda \hat{J}$ isexactly the same radial equation. The common factor $c_n$ cancels.

For two regular solutions with the same $n$ and different multipliers, multiply the equation for $f_1$ by $f_2$, multiply that for $f_2$ by $f_1$, subtract and integrate. The result isAt $a$, both functions vanish; at zero, regularity makes the boundary term tend to zero. Since the multipliers are distinct,This is weighted Sturm–Liouville orthogonality with weight $r$.

Finally, multiply one radial equation by $f$ and integrate by parts. The same endpoint conditions yieldMultiplication by $c_n$ provesFor any nonzero admissible solution the energy on the left before subtraction is positive, so its multiplier is the positive Rayleigh quotient $\lambda =\hat{I}/\hat{J}$.