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Let $f(t) = \int_0^t \ln{(s^2+t^2)} ds$, how can I find the derivative $\frac{df(t)}{dt}$?

The function $\,\int_0^t \ln{(s^2+t^2)} ds$ is defined to be continuous in $s^2+t^2 > 0$ and $ s^2+t^2 \ne 0$, then by the Fundamental Theorem of Integral Calculus, we have $$ f´(t) = (\int_0^t \ln{(s^2+t^2)} ds)´= f(t) - f(0) = \ln{(s^2+t^2)} + \ln{(s^2)}$$

Is that correct? can I apply the same procedure to this function $g(t) = \int_0^t \frac{e^{st}}{s} ds$ ? Please help. Thanks.

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4 Answers 4

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That is not correct. The integrand is itself a function of $t$. And clearly, the answer should be independent of $s$.

Check here: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

Hence, $$f'(t)=\ln(t^2+t^2)+\int_0^t \frac{2t}{s^2+t^2}\,ds=\ln(2t^2)+\frac{\pi}{2}$$

Similarly, $$g'(t)=\frac{e^{t^2}}{t}+\int_0^t e^{st}\,ds=\frac{e^{t^2}}{t}+\frac{e^{t^2}-1}{t}=\frac{2e^{t^2}-1}{t}$$

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    $\begingroup$ Your last term should be $\pi/2$ since $s=t$ as you wrote it just before. $\endgroup$ Jun 22, 2014 at 7:47
  • $\begingroup$ @ClaudeLeibovici: Oops, thanks! :P $\endgroup$ Jun 22, 2014 at 7:49
  • $\begingroup$ Sorry for my dumbs questions but you integrate directly f ? or applied the Fundamental Theorem, really thanx for your patient .... $\endgroup$ Jun 22, 2014 at 7:52
  • $\begingroup$ @Knight: I applied the Fundamental theorem of Calculus. Click the link I posted. And the integral is very common. What is the derivative of arctan(x)? $\endgroup$ Jun 22, 2014 at 7:55
  • $\begingroup$ ok, really thanx. $\endgroup$ Jun 22, 2014 at 7:57
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A non-orthodoxical approach.

Let $G(s,t)$ be a primitive of the integrand $g(s,t)$ (antiderivative on $s$): it is such that $\frac{\partial G}{\partial s}$ is the integrand.

The expression to be derived is thus $G(s,t)|_{s=0}^t=G(t,t)-G(0,t)$. When deriving with respect to $t$, we need to use the chain rule: $$\frac{dG(t,t)}{dt}=\frac{\partial G}{\partial s}(t,t)\frac{dt}{dt}+\frac{\partial G}{\partial t}(t,t)\frac{dt}{dt}=g(t,t)+\frac{\partial G}{\partial t}(t,t),$$ $$\frac{G(0,t)}{dt}=\frac{\partial G}{\partial s}(0,t)\frac{d0}{dt}+\frac{\partial G}{\partial t}(0,t)\frac{dt}{dt}=\frac{\partial G}{\partial t}(0,t).$$ We are still missing the expression of $\frac{\partial G}{\partial t}$. $G(s,t)$ being a primitive (on $s$) of $g(s,t)$, $\frac{\partial G}{\partial t}(s,t)$ must be a primitive (on $s$) of $\frac{\partial g}{\partial t}(s,t)$. So $$\frac{\partial G}{\partial t}(s,t)=\int \frac{2t}{s^2+t^2}ds=2\arctan\frac st+C.$$

Putting it all together, the answer is $$g(t,t)+2\arctan1-2\arctan0=\ln(2t^2)+\frac{\pi}2$$

For the second case, $$\frac{\partial G}{\partial t}(s,t)=\int{e^{st}}ds=\frac{e^{st}}t+C.$$

and the answer, $$\frac{e^{t^2}}t+\frac{e^{t^2}}t-\frac10.$$

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Hint: using the Leibniz integral rule,

$$\begin{align} \frac{d}{dt}f(t)&=\ln{(s^2+t^2)}\bigg{|}_{s=t}\frac{d}{dt}(t)-\ln{(s^2+t^2)}\bigg{|}_{s=0}\frac{d}{dt}(0)+\int_{0}^{t}\frac{\partial}{\partial t}\ln{(s^2+t^2)}\,\mathrm{d}s \end{align}$$

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The integration can be carried out explicitly:

$$\,\int_0^t \ln{(s^2+t^2)} ds=(1/2)t(-4 + \pi + 2\ln 2 + 4\ln t)=:f(t)$$

So

$$f'(t)=\pi/2 + \ln(2t^2)$$

And $\int_0^t \frac{e^{st}}{s} ds$ does not converge (at $s= 0^+$).

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