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Fixed Row Alignment.
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Lucian
Lucian
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No,Although there is no closed form. However in terms of elementary functions, $\displaystyle\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a~$$$\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a$$ should make for a decent lower limit. Though, depending on the ratio between a toand b, its value 

can be up to several $/$ many times smaller than that of the original. In any case, the asymptotic 

approximation seems tocan be the way to gosignificantly improved, perhaps by adding a second approximating function for the 

integrand, for small values of x, since the one mentioned above works better for larger values of 

the variable, as $x\to\infty$. Hope this helps.

No, there is no closed form. However, $\displaystyle\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a~$ should make for a decent lower limit. Though, depending on the ratio between a to b, its value can be up to several $/$ many times smaller than that of the original. In any case, asymptotic approximation seems to be the way to go, perhaps by adding a second approximating function for the integrand, for small values of x, since the one mentioned above works better for larger values of the variable, as $x\to\infty$. Hope this helps.

Although there is no closed form in terms of elementary functions, $$\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a$$ should make for a decent lower limit. Though, depending on the ratio between a and b, its value 

can be up to several $/$ many times smaller than that of the original. In any case, the asymptotic 

approximation can be significantly improved, by adding a second approximating function for the 

integrand, for small values of x, since the one mentioned above works better for larger values of 

the variable, as $x\to\infty$. Hope this helps.

deleted 123 characters in body
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Lucian
Lucian
  • 655
  • 1
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  • 22

No No, there is nono closed form. However, $\displaystyle\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a~$ should make for a decent lower limit. Though, depending on the ratio between a to b, its value can be up to several $/$ many times smaller than that of the original. In any case, asymptotic approximation seems to be the way to go, perhaps by adding a second approximating function for the integrand, for small values of x, since the one mentioned above works better for larger values of the variable, as $x\to\infty$. Hope this helps.

No, there is no closed form. However, $\displaystyle\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a~$ should make for a decent lower limit. Though, depending on the ratio between a to b, its value can be up to several $/$ many times smaller than that of the original. In any case, asymptotic approximation seems to be the way to go, perhaps by adding a second approximating function for the integrand, for small values of x, since the one mentioned above works better for larger values of the variable, as $x\to\infty$. Hope this helps.

No, there is no closed form. However, $\displaystyle\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a~$ should make for a decent lower limit. Though, depending on the ratio between a to b, its value can be up to several $/$ many times smaller than that of the original. In any case, asymptotic approximation seems to be the way to go, perhaps by adding a second approximating function for the integrand, for small values of x, since the one mentioned above works better for larger values of the variable, as $x\to\infty$. Hope this helps.

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Lucian
Lucian
  • 655
  • 1
  • 7
  • 22

No, there is no closed form. However, $\displaystyle\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a~$ should make for a decent lower limit. Though, depending on the ratio between a to b, its value can be up to several $/$ many times smaller than that of the original. In any case, asymptotic approximation seems to be the way to go, perhaps by adding a second approximating function for the integrand, for small values of x, since the one mentioned above works better for larger values of the variable, as $x\to\infty$. Hope this helps.