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If $f(x) = \int^1_0 e^{xy+y^2}dy$, find $f'(0)$.

I understand that this is function defined by an integral, and $e^{y^{2}}$ does not integrate into an elementary function. So, I will need to take $f'(x)$ which yields:

$\int^1_0 ye^{xy+y^2}dy$

I am trying to integrate this, but I am failing. I take it I should use integration by parts, but I can't because I still have $e^{y^2}$ term. Any help?

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    Derive under the integration sign, this is smooth and it will be allowed.2012-05-16

3 Answers 3

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You have got f'(x), so let x be 0, and integrate by parts. Then you get the answer (e-1)/2.

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    Pan Yan, if you look at the other two anwers, you can see you are simply repeating in a summarized way what has been already been explained. It isn't useful to repeat what other questions say, but rather to add **extra** remarks, ideas or information. This question, as it is, is not useful for the OP.2012-05-17
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You are given

$f(x) =\int_0^1 \exp(xy+y^2)dy $

Differentiating gives

$f'(x) =\int_0^1 y\cdot \exp(xy+y^2)dy $

Since we need $f'(0)$ we might as well plug in $x=0$. This gives

$f'(0) =\int_0^1 y\cdot \exp (y^2)\cdot dy $

But this integral is quite striaghtforward,

$f'(0) =\int_0^1 y\cdot \exp (y^2)\cdot dy =\left. \frac 1 2 e^{y^2} \right|_0^1 = \frac 1 2 (e-1) $


By the OP's request:

$f'(x) =\int_0^1 y\cdot \exp(xy+y^2)dy $

We complete the square

$f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_0^1 {y\exp \left[ {{{\left( {y + \frac{x}{2}} \right)}^2}} \right]dy} $

We change variables

$\eqalign{ & y + \frac{x}{2} = u \cr & dy = du \cr} $

$\eqalign{ & f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\left( {u - \frac{x}{2}} \right)} \exp \left( {{u^2}} \right)du \cr & f'(x) = \exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} u \exp \left( {{u^2}} \right)du - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\exp \left( {{u^2}} \right)du} \cr} $

Let's focus on the first integral:

${I_1} = \int_{\frac{x}{2}}^{1 + \frac{x}{2}} u \exp \left( {{u^2}} \right)du = \left. {\frac{1}{2}\exp {u^2}} \right|_{x/2}^{1 + x/2} = \frac{1}{2}\exp \frac{{{x^2}}}{4}\left[ {\exp \left( {x + 1} \right) - 1} \right]$

So we have

$f'(x) = \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right] - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\int_{\frac{x}{2}}^{1 + \frac{x}{2}} {\exp \left( {{u^2}} \right)du} $

The other integral evaluates in terms of the error function

$\operatorname{erf} (x) = \frac{2}{{\sqrt \pi }}\int_0^x {{e^{ - {t^2}}}} dt.$

with a change or variables $u=-v$,

$\int_{ - \left( {1 + \frac{x}{2}} \right)}^{ - \frac{x}{2}} {\exp \left( { - {v^2}} \right)dv} = \frac{{\sqrt \pi }}{2}\left\{ {\operatorname{erf} \left( { - \frac{x}{2}} \right) - \operatorname{erf} \left( { - 1 - \frac{x}{2}} \right)} \right\}$

So the function is

$f'(x) = \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right] - \frac{x}{2}\exp \left( { - \frac{{{x^2}}}{4}} \right)\frac{{\sqrt \pi }}{2}\left\{ {\operatorname{erf} \left( { - \frac{x}{2}} \right) - \operatorname{erf} \left( { - 1 - \frac{x}{2}} \right)} \right\}$

For large values of $x$, you can neglect the last result (since the value will be smaller and smaller), and you can estimate $f'$ with

$f'(x) \approx \frac{1}{2}\left[ {\exp \left( {x + 1} \right) - 1} \right]$

Cheers.

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    @Donnie Any time!2012-05-17
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You're asked to calculate $f^{\prime}(0)$. You don't have to calculate $f^{\prime}(x)$ first.

\begin{align} f^{\prime}(0) = \int_0^1 y e^{y^2}dy \end{align}

Let $u = y^2$, $du = 2ydy$:

\begin{align} f^{\prime}(0) &= \frac{1}{2} \int_0^1 e^u du = \left. \frac{1}{2}e^u \right|_0^1 \\ &= \frac{1}{2}(e - 1) \end{align}

As for the general case of $f^\prime(x)$, the integral cannot be solved using elementary functions only. One will have to use the error function.