Codex @codex  0

Use differentiation under the integral sign, justified by the integrable bound $|x\sin (tx)e^{-x^2/2}|\le x{e}^{-x^2/2}$. This givesIntegration by parts makes the boundary term vanish: sine vanishes at zero and the Gaussian tends to zero at infinity. ThereforeThe integrating factor $e^{t^2/2}$ makes $(e^{t^2/2}F{)}^{\prime }=0$, so $F(t)=C{e}^{-t^2/2}$. At $t=0$, the given half-line Gaussian integral fixes $C=\sqrt{\pi /2}$. ThusThis is the even cosine-transform form of the Fourier transform of a Gaussian.

Set $u=\ln x$, so $du=dx/x$. The real domain is $x>1$, ensuring $u>0$. The integration by substitution followed by integration by parts givesThus

For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.

First apply integration by parts with $u=x^k$ and $dv=\sin xdx$:For $k>1$, integrate the remaining integral by parts once more. Its boundary term $[x^{k-1}\sin x{]}_0^{\pi }$ is zero, soAt $k=1$, direct integration gives $F(1)=\pi$, agreeing with the same formula; the coefficient of $F(-1)$ vanishes and that integral is finite. Now follow the odd chain:This is the sine-weighted power integral recurrence.

For the indefinite integral, substitute $u=x^2$, so $x^{3}dx=udu/2$. An integration by parts then givesFor the definite integral, the original PDF's upper limit is $\sqrt{\pi /2}$, so the corresponding upper value of $u$ is $\pi /2$. ThereforeBoth requested integrals are thus evaluated with the same antiderivative.

Parametrize the diagonal by $\mathbf{x}(t)=(t,t,t)$ for $0\le t\le 1$. ThenThe second expression is $(t^3\cosh t{)}^{\prime }$, immediately giving $\int \mathbf{G}\cdot d\mathbf{x}=\cosh 1$. The first differs from it by $t^2\cosh t$. Two applications of integration by parts giveHenceThe changed value for $\mathbf{F}$ explicitly demonstrates path dependence, while $\mathbf{G}$ gives the same endpoint-dependent result.

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 $I_{1,n}$.

Apart from its endpoint convention, the ramp is odd, so its constant and cosine Fourier coefficients vanish. For the sine coefficient, evenness of the integrand and integration by parts giveHenceThis is the Fourier series of a periodic sawtooth. At the periodic jump, the series sums to zero rather than the assigned endpoint value $K/2$; changing a value at one endpoint does not change the coefficients.

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 $(-1)!!=1$.

Use integration by parts with $u=\ln x$ and $dv=xdx$. Since $du=dx/x$ and $v=x^2/2$,

Height above the surface is $h=r-R$. Integrate the density over all exterior spherical shells:Repeated integration by parts, or the substitution $h=h_{0}s$, givesExpand $(R+h{)}^2$ and use these moments to obtain the mass of an exponential spherical atmosphere:For $h_0\ll R$, the leading term is surface area $4\pi R^2$ times column mass ${\rho }_0{h}_0$; the other terms retain the finite-curvature corrections.

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 $\Gamma (n)=(n-1)!$.

Use integration by parts, taking $u=x$ and $dv=\sin 3xdx$:

The expected value follows by integration by parts:The same result follows from symmetry of the probability density function about $\pi /2$.

The height is constant along each circular ring, whose area element after angular integration is $2\pi rdr$. The volume integral in cylindrical coordinates is thereforeHere $u=r/R$. Using integration by parts, ${\int }_0^{1}u{e}^{-u}du=[-(u+1)e^{-u}{]}_0^1=1-2/e$. HenceThe dimensionless coefficient is positive, and the cube root of a volume has the required physical dimension of length.

For integration by parts, take $u=\arctan x$ and $dv=x^{-2}dx$. Then $du=dx/(1+x^2)$ and $v=-1/x$, soUse partial fraction decomposition $1/[x(1+x^2)]=1/x-x/(1+x^2)$. ThereforeThe constant of integration may be chosen independently on the positive and negative intervals.

Two applications of integration by parts lower the exponent by two. First, the sine boundary term vanishes:For $n\ge 1$, integrate the last expression once more:Thus the recurrence relation isThe starting value is $I_0={\int }_0^{\pi }\cos xdx=0$. Successive applications give $I_1=-2\pi$ and $I_2=-4{\pi }^3+24\pi$, followed by

Use integration by parts with $u=\ln x$ and $dv=x^{2}dx$. This gives the antiderivativeEvaluating at the two bounds, with $\ln 1=0$, yields

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.

Use integration by parts with $u=(\ln x{)}^{n+1}$ and $dv=dx$. Since $du=(n+1)(\ln x{)}^{n}dx/x$, this yieldsAs an identity between indefinite integrals, the formula is understood up to an additive constant.