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Let the real-valued function $\phi:\mathbb{R}\to\mathbb{R}$ be defined by $$\phi(t)=\int_0^1e^{\sqrt{x^2+t^2}}\,\mathrm{d}x,$$ it can then be shown that $\phi$ is continuous and differentiable.

I wish to find its derivative, in particular at $\mathbf{t = 0}$. I believe, since the integration is independent of $t$, that I can differentiate under the integration sign (correct me if I'm wrong) hence $$\phi'(t)=\frac{\mathrm{d}}{\mathrm{d}t}\int_0^1e^{\sqrt{x^2+t^2}}\,\mathrm{d}x = \int_0^1 \frac{\mathrm{d}}{\mathrm{d}t}e^{\sqrt{x^2+t^2}}\,\mathrm{d}x = \int_0^1 \frac{te^{\sqrt{x^2+t^2}}}{\sqrt{x^2+t^2}}\,\mathrm{d}x.$$ Naïvely, it then looks like $\phi'(0) = 0$, since it is $0$ times something. However, as $t\to0$ we actually get the indeterminate form $0\cdot \infty$ since the integrand becomes infinite, so we cannot say that $\phi'(0) = 0$.

I tried various substitutions, but I'm not sure if this is the right approach. Perhaps, it is necessary to use dominated convergence theorem or something of that sort?

Thanks for any help.

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    $\begingroup$ is this not an even function? so the derivative at $t = 0$ should be zero. $\endgroup$ – abel Apr 19 '15 at 1:00
  • $\begingroup$ True, if you know $\phi'(0)$ exists. $\endgroup$ – zhw. Apr 23 '15 at 18:18
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The differentiation under the integration sign is indeed a bit tricky. In general, we need the function $f$ that is being integrated (in this case $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R},\;f(x,t):=e^{\sqrt{x^2+t^2}}$) to be partially differentiable with respect to the "variable not involved in the integration", and that the partial derivative is continuous.

In your question, we have that for every $\epsilon>0$, $f$ is continuously partially differentiable with respect to the second variable when we restrict to $[\epsilon,2]\times \mathbb{R}$ (since then we avoid the singularity at $(0,0)$ ). So we could say that: $$\frac{d}{dt}\left(\int_{\epsilon}^{1}e^{\sqrt{x^2+t^2}}dx\right)=\int_{\epsilon}^{1}\frac{d}{dt}\left(e^{\sqrt{x^2+t^2}}\right)dx=\int_{\epsilon}^{1}\frac{te^{\sqrt{x^2+t^2}}}{\sqrt{x^2+t^2}}dx $$ Note that evaluating in $t=0$ gives $0$ without any problems.

Now under the assumption that $\phi'(0)$ exists, we find: $$\phi'(0)=\left.\lim_{\epsilon\to 0}\int_{\epsilon}^{1}\frac{te^{\sqrt{x^2+t^2}}}{\sqrt{x^2+t^2}}dx\right|_{t=0}=\lim_{\epsilon\to 0}\int_{\epsilon}^1 0\;dt=0 $$

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Your last integral equals

$$\int_0^1 \frac{e^{\sqrt{x^2+t^2}}}{\sqrt{(x/t)^2+1}}\,dx.$$

The integrands are uniformly bounded for small $t.$ They converge pointwise to $0.$ Hence $\phi'(t)\to 0$ as $t\to 0$ by the dominated convergence theorem. This implies $\phi'(0) = 0.$

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