# Double Integral Question

#### WrittenStars

I am trying to evaluate the following double integral, by changing the order of integration first.

cosh(x^2).dx.dy for 3y < or equal to: x < or equal to 3
and 0 < or equal to: y < or equal to 1

I changed the order of integration to get:

cosh(x^2).dy.dx for x/3 < or equal to: y < or equal to 1
and 0 < or equal to: x < or equal to 3

From here I used the fact that cosh(x^2) = 1/2e^(x^2) + 1/2e^(-x^2)

So I have:

1/2e^(x^2) + 1/2e^(-x^2).dy.dx
for x/3 < or equal to: y < or equal to 1
and 0 < or equal to: x < or equal to 3

From here I integrated and ended up with an answer of 0.

I'm not sure if the approach I used was correct. can i use cosh(x^2) = 1/2e^(x^2) + 1/2e^(-x^2) for the double integration.

#### HallsofIvy

MHF Helper
So you had $$\displaystyle \int_{y=0}^3\int_{x= 3y}^3 cosh(x^2)dx dy$$ and have changed it to
$$\displaystyle \int_{x= 0}^1\int_{y= x/3}^1 cosh(x^2) dy dx$$.

That conversion is valid.

However, when you integrate with respect to y, you will have
$$\displaystyle \int_{x=0}^1 \left[ y cosh(x^2)\right]_{y= x/3}^1 y cosh(x^2) dx$$
$$\displaystyle = \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx$$.

You certainly can make the substitution $$\displaystyle u= x^2$$ in the second integral because then du= 2x dx so $$\displaystyle \frac{1}{2}du= xdx$$. But I see no good way to integrate $$\displaystyle cosh(x^2)= \frac{e^{x^2}+ e^{-x^2}}{2}$$.

How did you do that?

• ques

#### WrittenStars

Thank you soo much for the reply and help. I basically just split it up into [e*(x^2)]/2 + [e*-(x^2)]/2, initially to try and integrate.

After using the way you suggested and evaluating it I got 0 again.

Not sure if I correctly went about the rest of the integration to get 0 though.

But from integrating cosh(x^2).dx for x between 0 and 3 i got:

First part: [sinh(x^2) / 2x] for x=0 and x=3, which led me to sinh(9)/6 - 0.

Second part: From integating (x/3)cosh(x^2).dx for x between 0 and 3 i got (1/6)sinh(9) - (1/6)sinh(0) = (1/6)sinh(9) - 0 = (1/6)sinh(9).

Hence subtracting the second integral from the first gives me 0 again.

Does that look right? Thanks again!

#### AllanCuz

So you had $$\displaystyle \int_{y=0}^3\int_{x= 3y}^3 cosh(x^2)dx dy$$ and have changed it to
$$\displaystyle \int_{x= 0}^1\int_{y= x/3}^1 cosh(x^2) dy dx$$.

That conversion is valid.

However, when you integrate with respect to y, you will have
$$\displaystyle \int_{x=0}^1 \left[ y cosh(x^2)\right]_{y= x/3}^1 y cosh(x^2) dx$$
$$\displaystyle = \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx$$.

You certainly can make the substitution $$\displaystyle u= x^2$$ in the second integral because then du= 2x dx so $$\displaystyle \frac{1}{2}du= xdx$$. But I see no good way to integrate $$\displaystyle cosh(x^2)= \frac{e^{x^2}+ e^{-x^2}}{2}$$.

How did you do that?
I think it might be sufficient to label $$\displaystyle \int \frac{e^{x^2}+ e^{-x^2}}{2}$$ as $$\displaystyle \frac{erf(-x) + erf(x)}{2}$$ where erf(x) is defined as $$\displaystyle erf(x) = \int e^{-x^2} dx$$ Note: http://mathworld.wolfram.com/Erf.html

Thank you soo much for the reply and help. I basically just split it up into [e*(x^2)]/2 + [e*-(x^2)]/2, initially to try and integrate.

After using the way you suggested and evaluating it I got 0 again.

Not sure if I correctly went about the rest of the integration to get 0 though.

But from integrating cosh(x^2).dx for x between 0 and 3 i got:

First part: [sinh(x^2) / 2x] for x=0 and x=3, which led me to sinh(9)/6 - 0.

Second part: From integating (x/3)cosh(x^2).dx for x between 0 and 3 i got (1/6)sinh(9) - (1/6)sinh(0) = (1/6)sinh(9) - 0 = (1/6)sinh(9).

Hence subtracting the second integral from the first gives me 0 again.

Does that look right? Thanks again!
I don't think is is correct. You're saying that

$$\displaystyle \int cosh(x^2) = \frac{sinh(x^2)}{2x}$$

But the derivative of $$\displaystyle \frac{sinh(x^2)}{2x}$$ is $$\displaystyle \frac{ cosh(x^2)x^{-1} - sinh(x^2)x^{-2} }{2}$$

In fact the parts of $$\displaystyle cosh(x^2)$$ is a well known error function in statistics, but does not have an elementary anti-derivative. I think it would follow that $$\displaystyle cosh(x^2)$$ does not either.

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• ques

#### HallsofIvy

MHF Helper
Well, you knew that $$\displaystyle \int e^{x^2}dx$$ and $$\displaystyle \int e^{-x^2}dx$$, individually, had no elementary anti-derivatives, didn't you?

It is not that they, and $$\displaystyle cosh(x^2)$$ and $$\displaystyle sinh(x^2)$$ "have no specific anti-derivatives". Of course, the have anti-derivatives but they cannot be written in terms of elementary functions.