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Assume $h$ is continuously differentiable on the interval concerned. For fixed $x>0$, take $u=h(t)$ and $dv=(x-t{)}^{-1/2}dt$, so $v=-2\sqrt{x-t}$. Integration by parts, interpreted first with an endpoint cutoff and then by a limit, givesThe upper endpoint term vanishes, and the lower endpoint supplies the explicit square-root term.

Write $\lambda =\ln ({\rho }^{-1})>0$. The probability density function normalizes when $1=B{\int }_0^{\infty }{e}^{-\lambda x}dx=B/\lambda$, so $B=\lambda$. Its survival function is $e^{-\lambda x}$ for $x\ge 0$; hence, for $x>y\ge 0$,Differentiating this conditional tail givesThe value assigned at the single endpoint does not affect any probability. The memorylessness of the exponential distribution says that $X-y$, conditional on $X>y$, is again an exponential distribution of rate $\lambda$. Its first two moments are $1/\lambda$ and $2/{\lambda }^2$, found by integration by parts, so translation leaves the variance unchanged:

Use the real Fourier series convention $f(x)\sim a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$ for period two. The Fourier coefficients areThe last two integrals follow by integration by parts. Thus the Fourier series of a half-wave rectified ramp isOn $(-1,1)$, $f(x)=|x|+x$. Its even part is $f_e(x)=|x|$ and its odd part is $f_o(x)=x$, so the cosine and sine terms separate:At a jump, the Fourier series converges to the average of the one-sided limits. Thus the first series for $f$ takes the value $1$ at odd integers, while the Fourier series of a periodic sawtooth for $f_o$ takes the value $0$ there. The even function has a continuous periodic extension.

At $x=0$, the formula for $f_e$ gives $0=1/2-(4/{\pi }^2){\sum }_{r\ge 0}(2r+1{)}^{-2}$. At $x=1/2$, the formula for $f_o$ has only odd-index terms and gives $1/2=(2/\pi ){\sum }_{r\ge 0}(-1{)}^r/(2r+1)$. Hence

Integration by parts with $u=x$ and $dv=e^{x}dx$ gives

The periodic ramp is an odd function, so $a_0=a_n=0$. Integration by parts givesThus the Fourier series of a periodic sawtooth isThe periodic function is smooth at $x=\pi /2$, so the Dirichlet-Jordan convergence theorem permits evaluation there. Even-index terms vanish and odd-index sines alternate in sign, giving

For $p>0$, integration by parts gives the Gamma function recurrence:The endpoint terms vanish because $x^p\to 0$ at zero and the exponential dominates the power at infinity. Also $\Gamma (1)={\int }_0^{\infty }{e}^{-x}dx=1$. Iterating for nonnegative integers $n$ therefore yieldsNegative integers are outside this integral definition and factorial conclusion.

Use the real Fourier series convention $g=a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$ for period two. Since $g$ is an even function, $b_n=0$, and $a_0={\int }_{-1}^1{x}^{2}dx=2/3$. Twice applying integration by parts, with $k=n\pi$, givesHenceThe periodic extension is continuous at its endpoints, so the series takes the specified endpoint values as well as the interior values.

Take $\sigma >0$ and substitute $u=(x-\mu )/\sigma$. The density becomes the standard Gaussian distribution density $\phi (u)=e^{-u^2/2}/\sqrt{2\pi }$. Its normalization follows from the supplied Gaussian integral after scaling by $\sqrt{2}$. Since $u\phi (u)$ is an integrable odd function, its integral is zero, and therefore $\mathbb{E}X=\mu$.

For the second moment, ${\phi }^{\prime }=-u\phi$. Integration by parts givesThe endpoint term vanishes by Gaussian decay. Consequently

For $x>0$, use integration by parts with $u=\ln x$ and $dv=dx$. ThenDifferentiating the answer checks cancellation of the extra constant term from $(x\ln x{)}^{\prime }$.

The integrand is an even function, since both $x$ and $\sin x$ are odd functions. Using integration by parts,The endpoint cosine terms vanish, leaving the sine contribution.

Repeated integration by parts gives, for every nonnegative integer $j$,With $A={\lambda }^2$, the first two moments are $\mathbb{E}[X]={\lambda }^2(2!/{\lambda }^3)=2/\lambda$ and $\mathbb{E}[X^2]={\lambda }^2(3!/{\lambda }^4)=6/{\lambda }^2$. Subtracting the square of the mean givesIf $X$ carries units, $\lambda$ has inverse units of $X$, so the mean and standard deviation have the units of $X$ and the variance has their square.

Normalize the probability density function on its nonnegative support:The integral follows from one integration by parts, with the boundary term zero since $\lambda >0$. Hence $A={\lambda }^2$. This is a gamma distribution of shape $2$ and rate $\lambda$.

Use integration by parts twice. First,The bracket containing $x\sin x$ vanishes, while the last definite integral is $2$. Therefore

Set $s=\sin u$, so $ds=\cos udu$, and write $J_n=I_{n,1}(\pi /2,0)$. The integration by substitution gives $J_n={\int }_0^1(1-s^3{)}^{n}ds$. For $n>0$, integration by parts yieldsBoth endpoint terms vanish. Use $s^3=1-(1-s^3)$ inside the remaining integral:ConsequentlyThis is the $p=3$ instance of the power-complement integral recurrence, with initial value $J_0=1$.

The Leibniz integral rule accounts for both the moving upper endpoint and the integrand's dependence on $a$:For the specified integrand this gives $I^{\prime }(a)=e^{-a^2}-{\int }_0^{a}x{e}^{-ax}dx$. When $a\ne 0$, integration by parts givessoFor the requested independent check, direct integration gives $I(a)=(1-e^{-a^2})/a$ at $a\ne 0$. Its derivative is $[2a^2{e}^{-a^2}-(1-e^{-a^2})]/a^2$, agreeing with the result above. At zero the apparent singularity is removable: $I(a)=a-a^3/2+O(a^5)$, so $\boxed{I(0)=0,I^{\prime }(0)=1}$.

Write the kernel as $K(s,t)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$, with $t>0$. Its partial derivatives areThus the heat kernel itself satisfies $K_t=K_{ss}$. The proposed convolution consequently givesThis verifies the heat equation. The parameter is positive time; the real integral formula is not defined at negative time.

The passage of derivatives through the integral needs justification. A sufficient condition is the Gaussian-weighted absolute integrability in heat-kernel convergence for Gaussian-integrable initial data; it holds, in particular, for data of polynomial growth. On compact positive-time intervals strictly below an integrable later time, the differentiated kernels have an integrable Gaussian bound, so differentiation under the integral sign applies.

Even if the PDF's assumed convergence is read as ordinary conditional convergence, one can justify the same calculation. Choose a later time $T>t$ and let $H(\xi )$ be a primitive of $\phi (\xi )e^{-{\xi }^2/(4T)}$. Convergence at spatial point zero and time $T$ makes $H$ bounded on both tails. Put $q(x,\xi ,t)=K(x-\xi ,t)e^{{\xi }^2/(4T)}$. This function, its spatial and time derivatives, and their $\xi$-derivatives decay as Gaussians when $t<T$. Integration by parts rewrites the convolution as $-\int H{q}_{\xi }d\xi$, with vanishing endpoint terms. Boundedness of $H$ now permits differentiation under this absolutely convergent integral; since $q_t=q_{xx}$, the same equation follows. Thus one need not silently equate conditional and absolute convergence.

The indicated integration by substitution givesPointwise, the argument of $\phi$ tends to $x$. Since the Gaussian integral is $\int e^{-{\eta }^2}d\eta =\sqrt{\pi }$, the expected limit is $\phi (x)$. To justify it for unbounded data, use the approximate identity property rather than assuming a uniform bound on $\phi$.

In the original variable, normalization givesFor $|\xi -x|<\delta$, continuity makes the difference smaller than any prescribed $\epsilon$ if $\delta$ is small enough, and the kernel mass is at most one. On the complement the Gaussian mass tends to zero; the data-weighted tail tends to zero by heat-kernel convergence for Gaussian-integrable initial data. This provesFor heat-kernel convergence from a conditionally convergent Gaussian integral, the same tail conclusion follows using the bounded primitive of $(\phi (\xi )-\phi (x))e^{-{\xi }^2/(4T)}$. The factor $q=K(x-\xi ,t)e^{{\xi }^2/(4T)}$ has its maximum at $\xi =xT/(T-t)$, approaching $x$. For sufficiently small $t$, it is monotone on each of the two tails outside $|\xi -x|<\delta$, and its values at the tail endpoints tend to zero exponentially. Integration by parts bounds each tail by a constant times its endpoint value, proving the same limit without an absolute-convergence assumption. The approach $t\downarrow 0$ is from positive time.

Over one period, the graph rises linearly from the limiting point $(-\pi ,0)$ to $(0,\pi )$, then drops to zero and stays there up to $\pi$. Repeat this pattern every $2\pi$. The given open endpoints do not fix values at the jumps, but changing individual point values does not change any Fourier coefficient.

Only the negative half-interval contributes. The constant coefficient isFor $n\ge 1$, integration by parts givesandConsequently the requested Fourier series isIt equals the given function on each open continuity interval. By the Dirichlet-Jordan convergence theorem, at $x=2k\pi$ it takes the jump midpoint $\pi /2$, while at $x=(2k+1)\pi$ it takes zero because the periodic one-sided limits agree there.

For real powers work on $x>0$. Apply integration by parts with $u=\ln x$ and $dv=x^{m}dx$, so $v=x^{m+1}/(m+1)$ when $m\ne -1$. Then

Retain the original PDF's full-line integration limits. For $n\ge 2$, write $x^n{e}^{-x^2}dx=-x^{n-1}d(e^{-x^2})/2$. Integration by parts givesGaussian decay dominates every power, so the endpoint term vanishes, soThis is the full-line Gaussian moment recurrence; its initial values differ from the half-line version.