# Integral over a finite Domain involving the Dirac-delta Function

In the formulation of the partial differential heat equation

$$\frac{\partial \theta}{\partial t}=\frac{\partial^2 \theta}{\partial x^2}, \hspace{1 cm}0\le x \le D$$

there is an incompatibility between the initial condition

$$t=0:\hspace{1 cm} \theta=0 \hspace{1 cm} 0\le x \le D$$

and the boundary condition at $$x=0$$

$$x=0:\hspace{1 cm} \theta=1 \hspace{1 cm} t\ge 0$$

that renders the analytical Fourier Series solution invalid at very small times from the initial condition. This incompatibility can be resolved by posing the initial condition as

$$t=0:\hspace{1 cm} \theta=H(-x)$$

where $$H(x)$$ is the Unit Heaviside Step Function. Plugging in this initial condition into the Fourier Series solution and skipping details, I end up having to evaluate the following integral

$$\int_0^D H(-x)\cos\lambda x dx$$

which following integration by parts and noting that the derivative of the Heaviside step Function is the Dirac-Delta function, reduces to

$$\frac{1}{\lambda}\int_0^D \delta(x)\sin\lambda x dx$$

and this is where I'm stuck. I do know that

$$\int_{-\infty}^\infty \delta(x)f(x) dx=f(0)$$

but I don't think that this property would be preserved if the integral is broken up into sums of integrals from $$-\infty$$ to $$0_-$$, $$0_+$$ to $$D_-$$ and $$D_+$$ to $$\infty$$.

Any suggestions on how to proceed would be deeply appreciated. Thanks

• well, $H(-x)=0$ for $x>0$, so the original integral is zero, assuming that it is an integral of Lebesgue – Masacroso Apr 20 '19 at 20:15
• Why care about the corners? – md2perpe Apr 20 '19 at 21:29
• Well, because the very short term solutions are bogus and it takes a while for the correct transient profile to evolve. – Sharat V Chandrasekhar Apr 21 '19 at 13:32
• @Masacroso, it's actually posed above as a Riemann integral, but with a redefinition of the measure, it could be cast as a Lebesque integral. – Sharat V Chandrasekhar Apr 21 '19 at 13:39
• @SharatVChandrasekhar ok, but the answer is the same, because $H(-x)=0$ for $x>0$, so the integral is zero. You are mixing things here, note that the Dirac delta cannot be stated inside a integral of Riemann because it is not a function, at most a measure or a distribution – Masacroso Apr 21 '19 at 13:41

Yes it is fine to say that $$\frac{1}{\lambda}\int_0^D \delta(x)\sin(\lambda x) = \sin(0) = 0$$.
You can think of the dirac delta as just being an infinite mass on the point $$0$$. Any other parts of the integral become irrelevant.
Also, you have to choose which condition to use at $$t=0, x=0$$ surely.