# [SOLVED]Tricky complex integral

#### dwsmith

##### Well-known member
$$\int_{\gamma}ze^{z^2}dz$$

$\gamma(t) = 2t + i -2ti$, for $0\leq t\leq 1$.

$\int_{\gamma} f(\gamma(t))\gamma'(t)dt$

But

$\int_{\gamma}ze^{z^2}dz \Rightarrow \frac{1}{2}\int e^wdw$

So then I would be solving

$$\frac{1}{2}\int\exp(4t-1+4ti-8t^2i)(4+4i-16ti)dw$$

Correct? And how would I find the appropriate bounds for this integral or would it still be 0 and 1?

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#### Opalg

##### MHB Oldtimer
Staff member
$$\int_{\gamma}ze^{z^2}dz$$

$\gamma(t) = 2t + i -2ti$, for $0\leq t\leq 1$.

$\int_{\gamma} f(\gamma(t))\gamma'(t)dt$

But

$\int_{\gamma}ze^{z^2}dz \Rightarrow \frac{1}{2}\int e^wdw$

So then I would be solving

$$\frac{1}{2}\int\exp(4t-1+4ti-8t^i)(4+4i-16ti)dw$$

Correct? And how would I find the appropriate bounds for this integral or would it still be 0 and 1?
As you correctly say, $\int ze^{z^2}dz = \frac{1}{2}\int e^wdw = \frac12e^w = \frac12e^{z^2}$. In other words, your function has an indefinite integral. The fundamental theorem of integration applies, and says that the value of the integral along the path $\gamma$ is the difference between the values of the indefinite integral at the endpoints of $\gamma$.

The endpoints of $\gamma$ are $\gamma(0) = i$ and $\gamma(1) = 2-i$. So the integral is equal to $\frac12\left(e^{(2-i)^2} - e^{i^2}\right)$.

• dwsmith

#### dwsmith

##### Well-known member
As you correctly say, $\int ze^{z^2}dz = \frac{1}{2}\int e^wdw = \frac12e^w = \frac12e^{z^2}$. In other words, your function has an indefinite integral. The fundamental theorem of integration applies, and says that the value of the integral along the path $\gamma$ is the difference between the values of the indefinite integral at the endpoints of $\gamma$.

The endpoints of $\gamma$ are $\gamma(0) = i$ and $\gamma(1) = 2-i$. So the integral is equal to $\frac12\left(e^{(2-i)^2} - e^{i^2}\right)$.
So you are evaluating this integral
$$\frac{1}{2}\int_{i}^{2-i}\exp\left(4t-14ti-8t^2i\right)(2-2i)dw$$

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#### Random Variable

##### Well-known member
MHB Math Helper
So you are evaluating this integral
$$\frac{1}{2}\int_{i}^{2-i}\exp\left(4t-14ti-8t^2i\right)(2-2i)dw$$

No, he's evaluating $\displaystyle \int_{i}^{2-i} z e^{z^{2}} \ dz$. The path taken doesn't matter.

I'll state the theorem directly from my old textbook.

Let $f$ be analytic in a simply connected domain D. If $z_{0}$ and $z_{1}$ are any two points in D joined by a contour $C$, then $\displaystyle \int_{C} f(z) \ dz = \int_{z_{0}}^{z_{1}}f(z) \ dz= F(z_{1})-F(z_{0})$ where $F$ is an antiderivative of $f$ in $D$.

$f(z) = ze^{z^{2}}$ is an entire function. It's analytic everywhere. So the above applies.

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• Prove It

#### dwsmith

##### Well-known member
No, he's evaluating $\displaystyle \int_{i}^{2-i} z e^{z^{2}} \ dz$. The path taken doesn't matter.

I'll state the theorem directly from my old textbook.

Let $f$ be analytic in a simply connected domain D. If $z_{0}$ and $z_{1}$ are any two points in D joined by a contour $C$, then $\displaystyle \int_{C} f(z) \ dz = \int_{z_{0}}^{z_{1}}f(z) \ dz= F(z_{1})-F(z_{0})$ where $F$ is an antiderivative of $f$ in $D$.

$f(z) = ze^{z^{2}}$ is an entire function. It's analytic everywhere. So the above applies.
After I make the substitution, I have $\displaystyle\frac{1}{2}\int_i^{2-i}e^wdw$ what is $w =$ now?

#### Random Variable

##### Well-known member
MHB Math Helper
$\displaystyle F(z) = \int f(z) = \int ze^{z^{2}} \ dz = \frac{1}{2} e^{z^{2}} + C$

so $\displaystyle \int_{\gamma} f(z) \ dz = \int_{i}^{2-i} f(z) \ dz = F(2-i) - F(i) = \frac{1}{2}e^{(2-i)^{2}} - \frac{1}{2} e^{i^{2}}$ and then simplify if you want

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• dwsmith

#### dwsmith

##### Well-known member
$\displaystyle F(z) = \int f(z) = \int ze^{z^{2}} = \frac{1}{2} e^{z^{2}} + C$

so $\displaystyle \int_{\gamma} f(z) \ dz = \int_{i}^{2-i} f(z) \ dz = F(2-i) - F(i) = \frac{1}{2}e^{(2-i)^{2}} - \frac{1}{2} e^{i^{2}}$ and then simply if you want
That is what I wasn't understanding.