Finding $f'(x)$ when $f(x)=\int^1_0 e^{xy+y^2}dy$

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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You want to calculate $f^{\prime}(0)$, so plug $x = 0$ and it will be an easy integral. – Ayman Hourieh May 16 '12 at 23:22
Derive under the integration sign, this is smooth and it will be allowed. – checkmath May 16 '12 at 23:27
Thank you Ayman – Donnie May 16 '12 at 23:33

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.

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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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If you put $x = 0$ in immediately after differentiating, you'll be able to skip a few steps... (edit: looks like you've done this now) – Ben Millwood May 16 '12 at 23:31
@benmachine I was thinking about finding the general $f'$ but then I changed my mind and that's why that mid step was kind of stranded there. – Pedro Tamaroff May 16 '12 at 23:31
@Peter, how would one go about finding the general f'? Is it possible? Thanks! – Donnie May 16 '12 at 23:33
@Donnie There is no elementary closed form of $f'$, but you can get a good approximation for large values of $x$. – Pedro Tamaroff May 16 '12 at 23:49
@PeterTamaroff Thank you very much. This is quite interesting. – Donnie May 17 '12 at 0:08

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. – Pedro Tamaroff May 17 '12 at 17:03