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I've recently come across a question that has completely stumped me, as follows:

Given a function $f(x)$ such that

$$ f(x) = \begin{cases} \frac{a}{x-1} [3 \sin (x-1) - 2 \tan (\ln x)], & \text{ if } 1/2 \lt x \leq 1 \\ b, & \text{ if } x=1 \\ \int ^{x^2} _{4(x-1)} e^{x + [\ln(t+1)]^c} \,dt, & \text{ if } x \gt 1. \end{cases} $$

Given that $f$ is differentiable, find the values of $a, b \text{ and }c.$

My immediate understanding is that since $f$ is differentiable, then it is also continuous for its entire domain. This means that taking the limits of each individual function as $x$ approaches 1 should yield the same result.

The first function was significantly easier to evaluate:

\begin{align} \lim\limits_{x \to 1^-} \frac{a}{x-1} [3 \sin (x-1) - 2 \tan (\ln x)] &= a \bigg(\lim\limits_{x \to 1^-}(3 \cos(x-1) - \frac{2 \sec^2(\ln x)}{x} \bigg) && \text{L'Hopital's Rule} \\ &= a \end{align}

And therefore $a=b$.

It's at the third function where I haven't the slightest idea of how to solve it. I've attempted, to no avail, integration by parts and converting the integral to a Riemann sum, but it feels as though I'm attempting to solve it through exhausting whatever techniques I have at hand.

I'd appreciate if someone could give an idea of how to understanding evaluating not just this particular integral, but integrals of similar form.

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  • $\begingroup$ @ganeshie8 nope, $c$ is a constant that I'm supposed to find. $\endgroup$ – fysh Aug 17 at 16:50
  • $\begingroup$ fysh instead of trying to integrate, take the derivative of that integral $\endgroup$ – AgentS Aug 17 at 17:08
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In the second integral, $x$ is a constant.

\begin{eqnarray} \int ^{x^2} _{4(x-1)} e^{x + [\ln(t+1)]^c} \,dt&=&e^x\int ^{x^2} _{4(x-1)}e^{\ln(t+1)^c}\,dt\\ &=&e^x\int ^{x^2} _{4(x-1)}(t+1)^c\,dt\\ &=&\frac{e^x}{c+1}[(x^2+1)^{c+1}-(4x-3)^{c+1}] \end{eqnarray} This must equal $b$ when $x=1$.

\begin{eqnarray} \left(\frac{2^{c+1}-1}{c+1}\right)\cdot e&=&b\\ \frac{e^{(c+1)\ln2}-1}{(c+1)\ln2}&=&\frac{b}{e\ln2}\tag{1} \end{eqnarray}

Substituting $u=(c+1)\ln2$ we get

$$ \frac{e^u-1}{u}=\frac{b}{e\ln2} $$

This can be solved using the Lambert $W$ function.

Wolfram alpha will give you the solution for equations of the form

$$ \frac{e^X-1}{X}=C $$ The solution is a bit daunting, but it will give you solution(s) for $c$ in terms of $a$, provided that $a$ lies in an appropriate range for the domain of $W$.

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